/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 14 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QReductionProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) QReductionProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPSizeChangeProof [EQUIVALENT, 0 ms] (18) YES (19) QDP (20) UsableRulesProof [EQUIVALENT, 0 ms] (21) QDP (22) QReductionProof [EQUIVALENT, 0 ms] (23) QDP (24) QDPSizeChangeProof [EQUIVALENT, 0 ms] (25) YES (26) QDP (27) UsableRulesProof [EQUIVALENT, 0 ms] (28) QDP (29) QReductionProof [EQUIVALENT, 0 ms] (30) QDP (31) QDPSizeChangeProof [EQUIVALENT, 0 ms] (32) YES (33) QDP (34) UsableRulesProof [EQUIVALENT, 0 ms] (35) QDP (36) QReductionProof [EQUIVALENT, 0 ms] (37) QDP (38) QDPSizeChangeProof [EQUIVALENT, 0 ms] (39) YES (40) QDP (41) UsableRulesProof [EQUIVALENT, 0 ms] (42) QDP (43) QReductionProof [EQUIVALENT, 0 ms] (44) QDP (45) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (46) QDP (47) PisEmptyProof [EQUIVALENT, 0 ms] (48) YES (49) QDP (50) UsableRulesProof [EQUIVALENT, 0 ms] (51) QDP (52) QReductionProof [EQUIVALENT, 0 ms] (53) QDP (54) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (55) QDP (56) PisEmptyProof [EQUIVALENT, 0 ms] (57) YES (58) QDP (59) UsableRulesProof [EQUIVALENT, 0 ms] (60) QDP (61) QDPOrderProof [EQUIVALENT, 28 ms] (62) QDP (63) QDPOrderProof [EQUIVALENT, 17 ms] (64) QDP (65) QDPOrderProof [EQUIVALENT, 0 ms] (66) QDP (67) QDPOrderProof [EQUIVALENT, 9 ms] (68) QDP (69) QDPOrderProof [EQUIVALENT, 11 ms] (70) QDP (71) QDPOrderProof [EQUIVALENT, 0 ms] (72) QDP (73) DependencyGraphProof [EQUIVALENT, 0 ms] (74) QDP (75) UsableRulesProof [EQUIVALENT, 0 ms] (76) QDP (77) QReductionProof [EQUIVALENT, 0 ms] (78) QDP (79) QDPSizeChangeProof [EQUIVALENT, 0 ms] (80) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: active(minus(0, Y)) -> mark(0) active(minus(s(X), s(Y))) -> mark(minus(X, Y)) active(geq(X, 0)) -> mark(true) active(geq(0, s(Y))) -> mark(false) active(geq(s(X), s(Y))) -> mark(geq(X, Y)) active(div(0, s(Y))) -> mark(0) active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) active(if(true, X, Y)) -> mark(X) active(if(false, X, Y)) -> mark(Y) mark(minus(X1, X2)) -> active(minus(X1, X2)) mark(0) -> active(0) mark(s(X)) -> active(s(mark(X))) mark(geq(X1, X2)) -> active(geq(X1, X2)) mark(true) -> active(true) mark(false) -> active(false) mark(div(X1, X2)) -> active(div(mark(X1), X2)) mark(if(X1, X2, X3)) -> active(if(mark(X1), X2, X3)) minus(mark(X1), X2) -> minus(X1, X2) minus(X1, mark(X2)) -> minus(X1, X2) minus(active(X1), X2) -> minus(X1, X2) minus(X1, active(X2)) -> minus(X1, X2) s(mark(X)) -> s(X) s(active(X)) -> s(X) geq(mark(X1), X2) -> geq(X1, X2) geq(X1, mark(X2)) -> geq(X1, X2) geq(active(X1), X2) -> geq(X1, X2) geq(X1, active(X2)) -> geq(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(X1, mark(X2)) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, mark(X2), X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVE(minus(0, Y)) -> MARK(0) ACTIVE(minus(s(X), s(Y))) -> MARK(minus(X, Y)) ACTIVE(minus(s(X), s(Y))) -> MINUS(X, Y) ACTIVE(geq(X, 0)) -> MARK(true) ACTIVE(geq(0, s(Y))) -> MARK(false) ACTIVE(geq(s(X), s(Y))) -> MARK(geq(X, Y)) ACTIVE(geq(s(X), s(Y))) -> GEQ(X, Y) ACTIVE(div(0, s(Y))) -> MARK(0) ACTIVE(div(s(X), s(Y))) -> MARK(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) ACTIVE(div(s(X), s(Y))) -> IF(geq(X, Y), s(div(minus(X, Y), s(Y))), 0) ACTIVE(div(s(X), s(Y))) -> GEQ(X, Y) ACTIVE(div(s(X), s(Y))) -> S(div(minus(X, Y), s(Y))) ACTIVE(div(s(X), s(Y))) -> DIV(minus(X, Y), s(Y)) ACTIVE(div(s(X), s(Y))) -> MINUS(X, Y) ACTIVE(if(true, X, Y)) -> MARK(X) ACTIVE(if(false, X, Y)) -> MARK(Y) MARK(minus(X1, X2)) -> ACTIVE(minus(X1, X2)) MARK(0) -> ACTIVE(0) MARK(s(X)) -> ACTIVE(s(mark(X))) MARK(s(X)) -> S(mark(X)) MARK(s(X)) -> MARK(X) MARK(geq(X1, X2)) -> ACTIVE(geq(X1, X2)) MARK(true) -> ACTIVE(true) MARK(false) -> ACTIVE(false) MARK(div(X1, X2)) -> ACTIVE(div(mark(X1), X2)) MARK(div(X1, X2)) -> DIV(mark(X1), X2) MARK(div(X1, X2)) -> MARK(X1) MARK(if(X1, X2, X3)) -> ACTIVE(if(mark(X1), X2, X3)) MARK(if(X1, X2, X3)) -> IF(mark(X1), X2, X3) MARK(if(X1, X2, X3)) -> MARK(X1) MINUS(mark(X1), X2) -> MINUS(X1, X2) MINUS(X1, mark(X2)) -> MINUS(X1, X2) MINUS(active(X1), X2) -> MINUS(X1, X2) MINUS(X1, active(X2)) -> MINUS(X1, X2) S(mark(X)) -> S(X) S(active(X)) -> S(X) GEQ(mark(X1), X2) -> GEQ(X1, X2) GEQ(X1, mark(X2)) -> GEQ(X1, X2) GEQ(active(X1), X2) -> GEQ(X1, X2) GEQ(X1, active(X2)) -> GEQ(X1, X2) DIV(mark(X1), X2) -> DIV(X1, X2) DIV(X1, mark(X2)) -> DIV(X1, X2) DIV(active(X1), X2) -> DIV(X1, X2) DIV(X1, active(X2)) -> DIV(X1, X2) IF(mark(X1), X2, X3) -> IF(X1, X2, X3) IF(X1, mark(X2), X3) -> IF(X1, X2, X3) IF(X1, X2, mark(X3)) -> IF(X1, X2, X3) IF(active(X1), X2, X3) -> IF(X1, X2, X3) IF(X1, active(X2), X3) -> IF(X1, X2, X3) IF(X1, X2, active(X3)) -> IF(X1, X2, X3) The TRS R consists of the following rules: active(minus(0, Y)) -> mark(0) active(minus(s(X), s(Y))) -> mark(minus(X, Y)) active(geq(X, 0)) -> mark(true) active(geq(0, s(Y))) -> mark(false) active(geq(s(X), s(Y))) -> mark(geq(X, Y)) active(div(0, s(Y))) -> mark(0) active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) active(if(true, X, Y)) -> mark(X) active(if(false, X, Y)) -> mark(Y) mark(minus(X1, X2)) -> active(minus(X1, X2)) mark(0) -> active(0) mark(s(X)) -> active(s(mark(X))) mark(geq(X1, X2)) -> active(geq(X1, X2)) mark(true) -> active(true) mark(false) -> active(false) mark(div(X1, X2)) -> active(div(mark(X1), X2)) mark(if(X1, X2, X3)) -> active(if(mark(X1), X2, X3)) minus(mark(X1), X2) -> minus(X1, X2) minus(X1, mark(X2)) -> minus(X1, X2) minus(active(X1), X2) -> minus(X1, X2) minus(X1, active(X2)) -> minus(X1, X2) s(mark(X)) -> s(X) s(active(X)) -> s(X) geq(mark(X1), X2) -> geq(X1, X2) geq(X1, mark(X2)) -> geq(X1, X2) geq(active(X1), X2) -> geq(X1, X2) geq(X1, active(X2)) -> geq(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(X1, mark(X2)) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, mark(X2), X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 8 SCCs with 17 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: IF(X1, mark(X2), X3) -> IF(X1, X2, X3) IF(mark(X1), X2, X3) -> IF(X1, X2, X3) IF(X1, X2, mark(X3)) -> IF(X1, X2, X3) IF(active(X1), X2, X3) -> IF(X1, X2, X3) IF(X1, active(X2), X3) -> IF(X1, X2, X3) IF(X1, X2, active(X3)) -> IF(X1, X2, X3) The TRS R consists of the following rules: active(minus(0, Y)) -> mark(0) active(minus(s(X), s(Y))) -> mark(minus(X, Y)) active(geq(X, 0)) -> mark(true) active(geq(0, s(Y))) -> mark(false) active(geq(s(X), s(Y))) -> mark(geq(X, Y)) active(div(0, s(Y))) -> mark(0) active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) active(if(true, X, Y)) -> mark(X) active(if(false, X, Y)) -> mark(Y) mark(minus(X1, X2)) -> active(minus(X1, X2)) mark(0) -> active(0) mark(s(X)) -> active(s(mark(X))) mark(geq(X1, X2)) -> active(geq(X1, X2)) mark(true) -> active(true) mark(false) -> active(false) mark(div(X1, X2)) -> active(div(mark(X1), X2)) mark(if(X1, X2, X3)) -> active(if(mark(X1), X2, X3)) minus(mark(X1), X2) -> minus(X1, X2) minus(X1, mark(X2)) -> minus(X1, X2) minus(active(X1), X2) -> minus(X1, X2) minus(X1, active(X2)) -> minus(X1, X2) s(mark(X)) -> s(X) s(active(X)) -> s(X) geq(mark(X1), X2) -> geq(X1, X2) geq(X1, mark(X2)) -> geq(X1, X2) geq(active(X1), X2) -> geq(X1, X2) geq(X1, active(X2)) -> geq(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(X1, mark(X2)) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, mark(X2), X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (6) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: IF(X1, mark(X2), X3) -> IF(X1, X2, X3) IF(mark(X1), X2, X3) -> IF(X1, X2, X3) IF(X1, X2, mark(X3)) -> IF(X1, X2, X3) IF(active(X1), X2, X3) -> IF(X1, X2, X3) IF(X1, active(X2), X3) -> IF(X1, X2, X3) IF(X1, X2, active(X3)) -> IF(X1, X2, X3) R is empty. The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: IF(X1, mark(X2), X3) -> IF(X1, X2, X3) IF(mark(X1), X2, X3) -> IF(X1, X2, X3) IF(X1, X2, mark(X3)) -> IF(X1, X2, X3) IF(active(X1), X2, X3) -> IF(X1, X2, X3) IF(X1, active(X2), X3) -> IF(X1, X2, X3) IF(X1, X2, active(X3)) -> IF(X1, X2, X3) R is empty. The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *IF(X1, mark(X2), X3) -> IF(X1, X2, X3) The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3 *IF(mark(X1), X2, X3) -> IF(X1, X2, X3) The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3 *IF(X1, X2, mark(X3)) -> IF(X1, X2, X3) The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3 *IF(active(X1), X2, X3) -> IF(X1, X2, X3) The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3 *IF(X1, active(X2), X3) -> IF(X1, X2, X3) The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3 *IF(X1, X2, active(X3)) -> IF(X1, X2, X3) The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3 ---------------------------------------- (11) YES ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: DIV(X1, mark(X2)) -> DIV(X1, X2) DIV(mark(X1), X2) -> DIV(X1, X2) DIV(active(X1), X2) -> DIV(X1, X2) DIV(X1, active(X2)) -> DIV(X1, X2) The TRS R consists of the following rules: active(minus(0, Y)) -> mark(0) active(minus(s(X), s(Y))) -> mark(minus(X, Y)) active(geq(X, 0)) -> mark(true) active(geq(0, s(Y))) -> mark(false) active(geq(s(X), s(Y))) -> mark(geq(X, Y)) active(div(0, s(Y))) -> mark(0) active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) active(if(true, X, Y)) -> mark(X) active(if(false, X, Y)) -> mark(Y) mark(minus(X1, X2)) -> active(minus(X1, X2)) mark(0) -> active(0) mark(s(X)) -> active(s(mark(X))) mark(geq(X1, X2)) -> active(geq(X1, X2)) mark(true) -> active(true) mark(false) -> active(false) mark(div(X1, X2)) -> active(div(mark(X1), X2)) mark(if(X1, X2, X3)) -> active(if(mark(X1), X2, X3)) minus(mark(X1), X2) -> minus(X1, X2) minus(X1, mark(X2)) -> minus(X1, X2) minus(active(X1), X2) -> minus(X1, X2) minus(X1, active(X2)) -> minus(X1, X2) s(mark(X)) -> s(X) s(active(X)) -> s(X) geq(mark(X1), X2) -> geq(X1, X2) geq(X1, mark(X2)) -> geq(X1, X2) geq(active(X1), X2) -> geq(X1, X2) geq(X1, active(X2)) -> geq(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(X1, mark(X2)) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, mark(X2), X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: DIV(X1, mark(X2)) -> DIV(X1, X2) DIV(mark(X1), X2) -> DIV(X1, X2) DIV(active(X1), X2) -> DIV(X1, X2) DIV(X1, active(X2)) -> DIV(X1, X2) R is empty. The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: DIV(X1, mark(X2)) -> DIV(X1, X2) DIV(mark(X1), X2) -> DIV(X1, X2) DIV(active(X1), X2) -> DIV(X1, X2) DIV(X1, active(X2)) -> DIV(X1, X2) R is empty. The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *DIV(X1, mark(X2)) -> DIV(X1, X2) The graph contains the following edges 1 >= 1, 2 > 2 *DIV(mark(X1), X2) -> DIV(X1, X2) The graph contains the following edges 1 > 1, 2 >= 2 *DIV(active(X1), X2) -> DIV(X1, X2) The graph contains the following edges 1 > 1, 2 >= 2 *DIV(X1, active(X2)) -> DIV(X1, X2) The graph contains the following edges 1 >= 1, 2 > 2 ---------------------------------------- (18) YES ---------------------------------------- (19) Obligation: Q DP problem: The TRS P consists of the following rules: GEQ(X1, mark(X2)) -> GEQ(X1, X2) GEQ(mark(X1), X2) -> GEQ(X1, X2) GEQ(active(X1), X2) -> GEQ(X1, X2) GEQ(X1, active(X2)) -> GEQ(X1, X2) The TRS R consists of the following rules: active(minus(0, Y)) -> mark(0) active(minus(s(X), s(Y))) -> mark(minus(X, Y)) active(geq(X, 0)) -> mark(true) active(geq(0, s(Y))) -> mark(false) active(geq(s(X), s(Y))) -> mark(geq(X, Y)) active(div(0, s(Y))) -> mark(0) active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) active(if(true, X, Y)) -> mark(X) active(if(false, X, Y)) -> mark(Y) mark(minus(X1, X2)) -> active(minus(X1, X2)) mark(0) -> active(0) mark(s(X)) -> active(s(mark(X))) mark(geq(X1, X2)) -> active(geq(X1, X2)) mark(true) -> active(true) mark(false) -> active(false) mark(div(X1, X2)) -> active(div(mark(X1), X2)) mark(if(X1, X2, X3)) -> active(if(mark(X1), X2, X3)) minus(mark(X1), X2) -> minus(X1, X2) minus(X1, mark(X2)) -> minus(X1, X2) minus(active(X1), X2) -> minus(X1, X2) minus(X1, active(X2)) -> minus(X1, X2) s(mark(X)) -> s(X) s(active(X)) -> s(X) geq(mark(X1), X2) -> geq(X1, X2) geq(X1, mark(X2)) -> geq(X1, X2) geq(active(X1), X2) -> geq(X1, X2) geq(X1, active(X2)) -> geq(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(X1, mark(X2)) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, mark(X2), X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (20) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (21) Obligation: Q DP problem: The TRS P consists of the following rules: GEQ(X1, mark(X2)) -> GEQ(X1, X2) GEQ(mark(X1), X2) -> GEQ(X1, X2) GEQ(active(X1), X2) -> GEQ(X1, X2) GEQ(X1, active(X2)) -> GEQ(X1, X2) R is empty. The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (22) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) ---------------------------------------- (23) Obligation: Q DP problem: The TRS P consists of the following rules: GEQ(X1, mark(X2)) -> GEQ(X1, X2) GEQ(mark(X1), X2) -> GEQ(X1, X2) GEQ(active(X1), X2) -> GEQ(X1, X2) GEQ(X1, active(X2)) -> GEQ(X1, X2) R is empty. The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (24) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *GEQ(X1, mark(X2)) -> GEQ(X1, X2) The graph contains the following edges 1 >= 1, 2 > 2 *GEQ(mark(X1), X2) -> GEQ(X1, X2) The graph contains the following edges 1 > 1, 2 >= 2 *GEQ(active(X1), X2) -> GEQ(X1, X2) The graph contains the following edges 1 > 1, 2 >= 2 *GEQ(X1, active(X2)) -> GEQ(X1, X2) The graph contains the following edges 1 >= 1, 2 > 2 ---------------------------------------- (25) YES ---------------------------------------- (26) Obligation: Q DP problem: The TRS P consists of the following rules: S(active(X)) -> S(X) S(mark(X)) -> S(X) The TRS R consists of the following rules: active(minus(0, Y)) -> mark(0) active(minus(s(X), s(Y))) -> mark(minus(X, Y)) active(geq(X, 0)) -> mark(true) active(geq(0, s(Y))) -> mark(false) active(geq(s(X), s(Y))) -> mark(geq(X, Y)) active(div(0, s(Y))) -> mark(0) active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) active(if(true, X, Y)) -> mark(X) active(if(false, X, Y)) -> mark(Y) mark(minus(X1, X2)) -> active(minus(X1, X2)) mark(0) -> active(0) mark(s(X)) -> active(s(mark(X))) mark(geq(X1, X2)) -> active(geq(X1, X2)) mark(true) -> active(true) mark(false) -> active(false) mark(div(X1, X2)) -> active(div(mark(X1), X2)) mark(if(X1, X2, X3)) -> active(if(mark(X1), X2, X3)) minus(mark(X1), X2) -> minus(X1, X2) minus(X1, mark(X2)) -> minus(X1, X2) minus(active(X1), X2) -> minus(X1, X2) minus(X1, active(X2)) -> minus(X1, X2) s(mark(X)) -> s(X) s(active(X)) -> s(X) geq(mark(X1), X2) -> geq(X1, X2) geq(X1, mark(X2)) -> geq(X1, X2) geq(active(X1), X2) -> geq(X1, X2) geq(X1, active(X2)) -> geq(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(X1, mark(X2)) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, mark(X2), X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (27) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (28) Obligation: Q DP problem: The TRS P consists of the following rules: S(active(X)) -> S(X) S(mark(X)) -> S(X) R is empty. The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (29) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) ---------------------------------------- (30) Obligation: Q DP problem: The TRS P consists of the following rules: S(active(X)) -> S(X) S(mark(X)) -> S(X) R is empty. The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (31) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *S(active(X)) -> S(X) The graph contains the following edges 1 > 1 *S(mark(X)) -> S(X) The graph contains the following edges 1 > 1 ---------------------------------------- (32) YES ---------------------------------------- (33) Obligation: Q DP problem: The TRS P consists of the following rules: MINUS(X1, mark(X2)) -> MINUS(X1, X2) MINUS(mark(X1), X2) -> MINUS(X1, X2) MINUS(active(X1), X2) -> MINUS(X1, X2) MINUS(X1, active(X2)) -> MINUS(X1, X2) The TRS R consists of the following rules: active(minus(0, Y)) -> mark(0) active(minus(s(X), s(Y))) -> mark(minus(X, Y)) active(geq(X, 0)) -> mark(true) active(geq(0, s(Y))) -> mark(false) active(geq(s(X), s(Y))) -> mark(geq(X, Y)) active(div(0, s(Y))) -> mark(0) active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) active(if(true, X, Y)) -> mark(X) active(if(false, X, Y)) -> mark(Y) mark(minus(X1, X2)) -> active(minus(X1, X2)) mark(0) -> active(0) mark(s(X)) -> active(s(mark(X))) mark(geq(X1, X2)) -> active(geq(X1, X2)) mark(true) -> active(true) mark(false) -> active(false) mark(div(X1, X2)) -> active(div(mark(X1), X2)) mark(if(X1, X2, X3)) -> active(if(mark(X1), X2, X3)) minus(mark(X1), X2) -> minus(X1, X2) minus(X1, mark(X2)) -> minus(X1, X2) minus(active(X1), X2) -> minus(X1, X2) minus(X1, active(X2)) -> minus(X1, X2) s(mark(X)) -> s(X) s(active(X)) -> s(X) geq(mark(X1), X2) -> geq(X1, X2) geq(X1, mark(X2)) -> geq(X1, X2) geq(active(X1), X2) -> geq(X1, X2) geq(X1, active(X2)) -> geq(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(X1, mark(X2)) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, mark(X2), X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (34) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (35) Obligation: Q DP problem: The TRS P consists of the following rules: MINUS(X1, mark(X2)) -> MINUS(X1, X2) MINUS(mark(X1), X2) -> MINUS(X1, X2) MINUS(active(X1), X2) -> MINUS(X1, X2) MINUS(X1, active(X2)) -> MINUS(X1, X2) R is empty. The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (36) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) ---------------------------------------- (37) Obligation: Q DP problem: The TRS P consists of the following rules: MINUS(X1, mark(X2)) -> MINUS(X1, X2) MINUS(mark(X1), X2) -> MINUS(X1, X2) MINUS(active(X1), X2) -> MINUS(X1, X2) MINUS(X1, active(X2)) -> MINUS(X1, X2) R is empty. The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (38) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *MINUS(X1, mark(X2)) -> MINUS(X1, X2) The graph contains the following edges 1 >= 1, 2 > 2 *MINUS(mark(X1), X2) -> MINUS(X1, X2) The graph contains the following edges 1 > 1, 2 >= 2 *MINUS(active(X1), X2) -> MINUS(X1, X2) The graph contains the following edges 1 > 1, 2 >= 2 *MINUS(X1, active(X2)) -> MINUS(X1, X2) The graph contains the following edges 1 >= 1, 2 > 2 ---------------------------------------- (39) YES ---------------------------------------- (40) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVE(geq(s(X), s(Y))) -> MARK(geq(X, Y)) MARK(geq(X1, X2)) -> ACTIVE(geq(X1, X2)) The TRS R consists of the following rules: active(minus(0, Y)) -> mark(0) active(minus(s(X), s(Y))) -> mark(minus(X, Y)) active(geq(X, 0)) -> mark(true) active(geq(0, s(Y))) -> mark(false) active(geq(s(X), s(Y))) -> mark(geq(X, Y)) active(div(0, s(Y))) -> mark(0) active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) active(if(true, X, Y)) -> mark(X) active(if(false, X, Y)) -> mark(Y) mark(minus(X1, X2)) -> active(minus(X1, X2)) mark(0) -> active(0) mark(s(X)) -> active(s(mark(X))) mark(geq(X1, X2)) -> active(geq(X1, X2)) mark(true) -> active(true) mark(false) -> active(false) mark(div(X1, X2)) -> active(div(mark(X1), X2)) mark(if(X1, X2, X3)) -> active(if(mark(X1), X2, X3)) minus(mark(X1), X2) -> minus(X1, X2) minus(X1, mark(X2)) -> minus(X1, X2) minus(active(X1), X2) -> minus(X1, X2) minus(X1, active(X2)) -> minus(X1, X2) s(mark(X)) -> s(X) s(active(X)) -> s(X) geq(mark(X1), X2) -> geq(X1, X2) geq(X1, mark(X2)) -> geq(X1, X2) geq(active(X1), X2) -> geq(X1, X2) geq(X1, active(X2)) -> geq(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(X1, mark(X2)) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, mark(X2), X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (41) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (42) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVE(geq(s(X), s(Y))) -> MARK(geq(X, Y)) MARK(geq(X1, X2)) -> ACTIVE(geq(X1, X2)) R is empty. The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (43) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) ---------------------------------------- (44) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVE(geq(s(X), s(Y))) -> MARK(geq(X, Y)) MARK(geq(X1, X2)) -> ACTIVE(geq(X1, X2)) R is empty. The set Q consists of the following terms: s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (45) UsableRulesReductionPairsProof (EQUIVALENT) By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well. The following dependency pairs can be deleted: ACTIVE(geq(s(X), s(Y))) -> MARK(geq(X, Y)) MARK(geq(X1, X2)) -> ACTIVE(geq(X1, X2)) No rules are removed from R. Used ordering: POLO with Polynomial interpretation [POLO]: POL(ACTIVE(x_1)) = x_1 POL(MARK(x_1)) = 2*x_1 POL(geq(x_1, x_2)) = 2 + 2*x_1 + 2*x_2 POL(s(x_1)) = 2 + 2*x_1 ---------------------------------------- (46) Obligation: Q DP problem: P is empty. R is empty. The set Q consists of the following terms: s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (47) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (48) YES ---------------------------------------- (49) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVE(minus(s(X), s(Y))) -> MARK(minus(X, Y)) MARK(minus(X1, X2)) -> ACTIVE(minus(X1, X2)) The TRS R consists of the following rules: active(minus(0, Y)) -> mark(0) active(minus(s(X), s(Y))) -> mark(minus(X, Y)) active(geq(X, 0)) -> mark(true) active(geq(0, s(Y))) -> mark(false) active(geq(s(X), s(Y))) -> mark(geq(X, Y)) active(div(0, s(Y))) -> mark(0) active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) active(if(true, X, Y)) -> mark(X) active(if(false, X, Y)) -> mark(Y) mark(minus(X1, X2)) -> active(minus(X1, X2)) mark(0) -> active(0) mark(s(X)) -> active(s(mark(X))) mark(geq(X1, X2)) -> active(geq(X1, X2)) mark(true) -> active(true) mark(false) -> active(false) mark(div(X1, X2)) -> active(div(mark(X1), X2)) mark(if(X1, X2, X3)) -> active(if(mark(X1), X2, X3)) minus(mark(X1), X2) -> minus(X1, X2) minus(X1, mark(X2)) -> minus(X1, X2) minus(active(X1), X2) -> minus(X1, X2) minus(X1, active(X2)) -> minus(X1, X2) s(mark(X)) -> s(X) s(active(X)) -> s(X) geq(mark(X1), X2) -> geq(X1, X2) geq(X1, mark(X2)) -> geq(X1, X2) geq(active(X1), X2) -> geq(X1, X2) geq(X1, active(X2)) -> geq(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(X1, mark(X2)) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, mark(X2), X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (50) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (51) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVE(minus(s(X), s(Y))) -> MARK(minus(X, Y)) MARK(minus(X1, X2)) -> ACTIVE(minus(X1, X2)) R is empty. The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (52) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) ---------------------------------------- (53) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVE(minus(s(X), s(Y))) -> MARK(minus(X, Y)) MARK(minus(X1, X2)) -> ACTIVE(minus(X1, X2)) R is empty. The set Q consists of the following terms: minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (54) UsableRulesReductionPairsProof (EQUIVALENT) By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well. The following dependency pairs can be deleted: ACTIVE(minus(s(X), s(Y))) -> MARK(minus(X, Y)) MARK(minus(X1, X2)) -> ACTIVE(minus(X1, X2)) No rules are removed from R. Used ordering: POLO with Polynomial interpretation [POLO]: POL(ACTIVE(x_1)) = x_1 POL(MARK(x_1)) = 2*x_1 POL(minus(x_1, x_2)) = 2 + 2*x_1 + 2*x_2 POL(s(x_1)) = 2 + 2*x_1 ---------------------------------------- (55) Obligation: Q DP problem: P is empty. R is empty. The set Q consists of the following terms: minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (56) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (57) YES ---------------------------------------- (58) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVE(div(s(X), s(Y))) -> MARK(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) MARK(if(X1, X2, X3)) -> ACTIVE(if(mark(X1), X2, X3)) ACTIVE(if(true, X, Y)) -> MARK(X) MARK(s(X)) -> ACTIVE(s(mark(X))) ACTIVE(if(false, X, Y)) -> MARK(Y) MARK(s(X)) -> MARK(X) MARK(div(X1, X2)) -> ACTIVE(div(mark(X1), X2)) MARK(div(X1, X2)) -> MARK(X1) MARK(if(X1, X2, X3)) -> MARK(X1) The TRS R consists of the following rules: active(minus(0, Y)) -> mark(0) active(minus(s(X), s(Y))) -> mark(minus(X, Y)) active(geq(X, 0)) -> mark(true) active(geq(0, s(Y))) -> mark(false) active(geq(s(X), s(Y))) -> mark(geq(X, Y)) active(div(0, s(Y))) -> mark(0) active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) active(if(true, X, Y)) -> mark(X) active(if(false, X, Y)) -> mark(Y) mark(minus(X1, X2)) -> active(minus(X1, X2)) mark(0) -> active(0) mark(s(X)) -> active(s(mark(X))) mark(geq(X1, X2)) -> active(geq(X1, X2)) mark(true) -> active(true) mark(false) -> active(false) mark(div(X1, X2)) -> active(div(mark(X1), X2)) mark(if(X1, X2, X3)) -> active(if(mark(X1), X2, X3)) minus(mark(X1), X2) -> minus(X1, X2) minus(X1, mark(X2)) -> minus(X1, X2) minus(active(X1), X2) -> minus(X1, X2) minus(X1, active(X2)) -> minus(X1, X2) s(mark(X)) -> s(X) s(active(X)) -> s(X) geq(mark(X1), X2) -> geq(X1, X2) geq(X1, mark(X2)) -> geq(X1, X2) geq(active(X1), X2) -> geq(X1, X2) geq(X1, active(X2)) -> geq(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(X1, mark(X2)) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, mark(X2), X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (59) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (60) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVE(div(s(X), s(Y))) -> MARK(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) MARK(if(X1, X2, X3)) -> ACTIVE(if(mark(X1), X2, X3)) ACTIVE(if(true, X, Y)) -> MARK(X) MARK(s(X)) -> ACTIVE(s(mark(X))) ACTIVE(if(false, X, Y)) -> MARK(Y) MARK(s(X)) -> MARK(X) MARK(div(X1, X2)) -> ACTIVE(div(mark(X1), X2)) MARK(div(X1, X2)) -> MARK(X1) MARK(if(X1, X2, X3)) -> MARK(X1) The TRS R consists of the following rules: active(minus(s(X), s(Y))) -> mark(minus(X, Y)) mark(minus(X1, X2)) -> active(minus(X1, X2)) mark(0) -> active(0) active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) mark(if(X1, X2, X3)) -> active(if(mark(X1), X2, X3)) active(if(true, X, Y)) -> mark(X) mark(s(X)) -> active(s(mark(X))) active(if(false, X, Y)) -> mark(Y) mark(div(X1, X2)) -> active(div(mark(X1), X2)) active(geq(s(X), s(Y))) -> mark(geq(X, Y)) mark(geq(X1, X2)) -> active(geq(X1, X2)) mark(true) -> active(true) mark(false) -> active(false) div(X1, mark(X2)) -> div(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) active(geq(X, 0)) -> mark(true) active(geq(0, s(Y))) -> mark(false) active(minus(0, Y)) -> mark(0) active(div(0, s(Y))) -> mark(0) if(X1, mark(X2), X3) -> if(X1, X2, X3) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) s(active(X)) -> s(X) s(mark(X)) -> s(X) The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (61) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. MARK(s(X)) -> ACTIVE(s(mark(X))) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(ACTIVE(x_1)) = x_1 POL(MARK(x_1)) = 1 POL(active(x_1)) = 0 POL(div(x_1, x_2)) = 1 POL(false) = 0 POL(geq(x_1, x_2)) = 0 POL(if(x_1, x_2, x_3)) = 1 POL(mark(x_1)) = 0 POL(minus(x_1, x_2)) = 0 POL(s(x_1)) = 0 POL(true) = 0 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: if(X1, mark(X2), X3) -> if(X1, X2, X3) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) s(active(X)) -> s(X) s(mark(X)) -> s(X) div(X1, mark(X2)) -> div(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) ---------------------------------------- (62) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVE(div(s(X), s(Y))) -> MARK(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) MARK(if(X1, X2, X3)) -> ACTIVE(if(mark(X1), X2, X3)) ACTIVE(if(true, X, Y)) -> MARK(X) ACTIVE(if(false, X, Y)) -> MARK(Y) MARK(s(X)) -> MARK(X) MARK(div(X1, X2)) -> ACTIVE(div(mark(X1), X2)) MARK(div(X1, X2)) -> MARK(X1) MARK(if(X1, X2, X3)) -> MARK(X1) The TRS R consists of the following rules: active(minus(s(X), s(Y))) -> mark(minus(X, Y)) mark(minus(X1, X2)) -> active(minus(X1, X2)) mark(0) -> active(0) active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) mark(if(X1, X2, X3)) -> active(if(mark(X1), X2, X3)) active(if(true, X, Y)) -> mark(X) mark(s(X)) -> active(s(mark(X))) active(if(false, X, Y)) -> mark(Y) mark(div(X1, X2)) -> active(div(mark(X1), X2)) active(geq(s(X), s(Y))) -> mark(geq(X, Y)) mark(geq(X1, X2)) -> active(geq(X1, X2)) mark(true) -> active(true) mark(false) -> active(false) div(X1, mark(X2)) -> div(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) active(geq(X, 0)) -> mark(true) active(geq(0, s(Y))) -> mark(false) active(minus(0, Y)) -> mark(0) active(div(0, s(Y))) -> mark(0) if(X1, mark(X2), X3) -> if(X1, X2, X3) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) s(active(X)) -> s(X) s(mark(X)) -> s(X) The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (63) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. MARK(div(X1, X2)) -> MARK(X1) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(ACTIVE(x_1)) = x_1 POL(MARK(x_1)) = x_1 POL(active(x_1)) = x_1 POL(div(x_1, x_2)) = 1 + x_1 POL(false) = 0 POL(geq(x_1, x_2)) = 0 POL(if(x_1, x_2, x_3)) = x_1 + x_2 + x_3 POL(mark(x_1)) = x_1 POL(minus(x_1, x_2)) = 0 POL(s(x_1)) = x_1 POL(true) = 0 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: active(minus(s(X), s(Y))) -> mark(minus(X, Y)) mark(minus(X1, X2)) -> active(minus(X1, X2)) mark(0) -> active(0) active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) mark(if(X1, X2, X3)) -> active(if(mark(X1), X2, X3)) active(if(true, X, Y)) -> mark(X) mark(s(X)) -> active(s(mark(X))) active(if(false, X, Y)) -> mark(Y) mark(div(X1, X2)) -> active(div(mark(X1), X2)) active(geq(s(X), s(Y))) -> mark(geq(X, Y)) mark(geq(X1, X2)) -> active(geq(X1, X2)) mark(true) -> active(true) mark(false) -> active(false) if(X1, mark(X2), X3) -> if(X1, X2, X3) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) div(X1, mark(X2)) -> div(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) active(minus(0, Y)) -> mark(0) active(geq(X, 0)) -> mark(true) active(geq(0, s(Y))) -> mark(false) active(div(0, s(Y))) -> mark(0) s(active(X)) -> s(X) s(mark(X)) -> s(X) ---------------------------------------- (64) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVE(div(s(X), s(Y))) -> MARK(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) MARK(if(X1, X2, X3)) -> ACTIVE(if(mark(X1), X2, X3)) ACTIVE(if(true, X, Y)) -> MARK(X) ACTIVE(if(false, X, Y)) -> MARK(Y) MARK(s(X)) -> MARK(X) MARK(div(X1, X2)) -> ACTIVE(div(mark(X1), X2)) MARK(if(X1, X2, X3)) -> MARK(X1) The TRS R consists of the following rules: active(minus(s(X), s(Y))) -> mark(minus(X, Y)) mark(minus(X1, X2)) -> active(minus(X1, X2)) mark(0) -> active(0) active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) mark(if(X1, X2, X3)) -> active(if(mark(X1), X2, X3)) active(if(true, X, Y)) -> mark(X) mark(s(X)) -> active(s(mark(X))) active(if(false, X, Y)) -> mark(Y) mark(div(X1, X2)) -> active(div(mark(X1), X2)) active(geq(s(X), s(Y))) -> mark(geq(X, Y)) mark(geq(X1, X2)) -> active(geq(X1, X2)) mark(true) -> active(true) mark(false) -> active(false) div(X1, mark(X2)) -> div(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) active(geq(X, 0)) -> mark(true) active(geq(0, s(Y))) -> mark(false) active(minus(0, Y)) -> mark(0) active(div(0, s(Y))) -> mark(0) if(X1, mark(X2), X3) -> if(X1, X2, X3) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) s(active(X)) -> s(X) s(mark(X)) -> s(X) The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (65) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. MARK(s(X)) -> MARK(X) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(ACTIVE(x_1)) = x_1 POL(MARK(x_1)) = x_1 POL(active(x_1)) = x_1 POL(div(x_1, x_2)) = x_1 POL(false) = 0 POL(geq(x_1, x_2)) = 0 POL(if(x_1, x_2, x_3)) = x_1 + x_2 + x_3 POL(mark(x_1)) = x_1 POL(minus(x_1, x_2)) = 0 POL(s(x_1)) = 1 + x_1 POL(true) = 0 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: active(minus(s(X), s(Y))) -> mark(minus(X, Y)) mark(minus(X1, X2)) -> active(minus(X1, X2)) mark(0) -> active(0) active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) mark(if(X1, X2, X3)) -> active(if(mark(X1), X2, X3)) active(if(true, X, Y)) -> mark(X) mark(s(X)) -> active(s(mark(X))) active(if(false, X, Y)) -> mark(Y) mark(div(X1, X2)) -> active(div(mark(X1), X2)) active(geq(s(X), s(Y))) -> mark(geq(X, Y)) mark(geq(X1, X2)) -> active(geq(X1, X2)) mark(true) -> active(true) mark(false) -> active(false) if(X1, mark(X2), X3) -> if(X1, X2, X3) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) div(X1, mark(X2)) -> div(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) active(minus(0, Y)) -> mark(0) active(geq(X, 0)) -> mark(true) active(geq(0, s(Y))) -> mark(false) active(div(0, s(Y))) -> mark(0) s(active(X)) -> s(X) s(mark(X)) -> s(X) ---------------------------------------- (66) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVE(div(s(X), s(Y))) -> MARK(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) MARK(if(X1, X2, X3)) -> ACTIVE(if(mark(X1), X2, X3)) ACTIVE(if(true, X, Y)) -> MARK(X) ACTIVE(if(false, X, Y)) -> MARK(Y) MARK(div(X1, X2)) -> ACTIVE(div(mark(X1), X2)) MARK(if(X1, X2, X3)) -> MARK(X1) The TRS R consists of the following rules: active(minus(s(X), s(Y))) -> mark(minus(X, Y)) mark(minus(X1, X2)) -> active(minus(X1, X2)) mark(0) -> active(0) active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) mark(if(X1, X2, X3)) -> active(if(mark(X1), X2, X3)) active(if(true, X, Y)) -> mark(X) mark(s(X)) -> active(s(mark(X))) active(if(false, X, Y)) -> mark(Y) mark(div(X1, X2)) -> active(div(mark(X1), X2)) active(geq(s(X), s(Y))) -> mark(geq(X, Y)) mark(geq(X1, X2)) -> active(geq(X1, X2)) mark(true) -> active(true) mark(false) -> active(false) div(X1, mark(X2)) -> div(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) active(geq(X, 0)) -> mark(true) active(geq(0, s(Y))) -> mark(false) active(minus(0, Y)) -> mark(0) active(div(0, s(Y))) -> mark(0) if(X1, mark(X2), X3) -> if(X1, X2, X3) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) s(active(X)) -> s(X) s(mark(X)) -> s(X) The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (67) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. ACTIVE(div(s(X), s(Y))) -> MARK(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(ACTIVE(x_1)) = x_1 POL(MARK(x_1)) = x_1 POL(active(x_1)) = x_1 POL(div(x_1, x_2)) = x_1 + x_2 POL(false) = 0 POL(geq(x_1, x_2)) = 0 POL(if(x_1, x_2, x_3)) = x_1 + x_2 + x_3 POL(mark(x_1)) = x_1 POL(minus(x_1, x_2)) = 0 POL(s(x_1)) = 1 POL(true) = 0 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: active(minus(s(X), s(Y))) -> mark(minus(X, Y)) mark(minus(X1, X2)) -> active(minus(X1, X2)) mark(0) -> active(0) active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) mark(if(X1, X2, X3)) -> active(if(mark(X1), X2, X3)) active(if(true, X, Y)) -> mark(X) mark(s(X)) -> active(s(mark(X))) active(if(false, X, Y)) -> mark(Y) mark(div(X1, X2)) -> active(div(mark(X1), X2)) active(geq(s(X), s(Y))) -> mark(geq(X, Y)) mark(geq(X1, X2)) -> active(geq(X1, X2)) mark(true) -> active(true) mark(false) -> active(false) if(X1, mark(X2), X3) -> if(X1, X2, X3) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) div(X1, mark(X2)) -> div(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) active(minus(0, Y)) -> mark(0) active(geq(X, 0)) -> mark(true) active(geq(0, s(Y))) -> mark(false) active(div(0, s(Y))) -> mark(0) s(active(X)) -> s(X) s(mark(X)) -> s(X) ---------------------------------------- (68) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(if(X1, X2, X3)) -> ACTIVE(if(mark(X1), X2, X3)) ACTIVE(if(true, X, Y)) -> MARK(X) ACTIVE(if(false, X, Y)) -> MARK(Y) MARK(div(X1, X2)) -> ACTIVE(div(mark(X1), X2)) MARK(if(X1, X2, X3)) -> MARK(X1) The TRS R consists of the following rules: active(minus(s(X), s(Y))) -> mark(minus(X, Y)) mark(minus(X1, X2)) -> active(minus(X1, X2)) mark(0) -> active(0) active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) mark(if(X1, X2, X3)) -> active(if(mark(X1), X2, X3)) active(if(true, X, Y)) -> mark(X) mark(s(X)) -> active(s(mark(X))) active(if(false, X, Y)) -> mark(Y) mark(div(X1, X2)) -> active(div(mark(X1), X2)) active(geq(s(X), s(Y))) -> mark(geq(X, Y)) mark(geq(X1, X2)) -> active(geq(X1, X2)) mark(true) -> active(true) mark(false) -> active(false) div(X1, mark(X2)) -> div(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) active(geq(X, 0)) -> mark(true) active(geq(0, s(Y))) -> mark(false) active(minus(0, Y)) -> mark(0) active(div(0, s(Y))) -> mark(0) if(X1, mark(X2), X3) -> if(X1, X2, X3) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) s(active(X)) -> s(X) s(mark(X)) -> s(X) The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (69) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. MARK(div(X1, X2)) -> ACTIVE(div(mark(X1), X2)) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(ACTIVE(x_1)) = x_1 POL(MARK(x_1)) = 1 POL(active(x_1)) = 0 POL(div(x_1, x_2)) = 0 POL(false) = 0 POL(geq(x_1, x_2)) = 0 POL(if(x_1, x_2, x_3)) = 1 POL(mark(x_1)) = 0 POL(minus(x_1, x_2)) = 0 POL(s(x_1)) = 0 POL(true) = 0 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: if(X1, mark(X2), X3) -> if(X1, X2, X3) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) div(X1, mark(X2)) -> div(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) ---------------------------------------- (70) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(if(X1, X2, X3)) -> ACTIVE(if(mark(X1), X2, X3)) ACTIVE(if(true, X, Y)) -> MARK(X) ACTIVE(if(false, X, Y)) -> MARK(Y) MARK(if(X1, X2, X3)) -> MARK(X1) The TRS R consists of the following rules: active(minus(s(X), s(Y))) -> mark(minus(X, Y)) mark(minus(X1, X2)) -> active(minus(X1, X2)) mark(0) -> active(0) active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) mark(if(X1, X2, X3)) -> active(if(mark(X1), X2, X3)) active(if(true, X, Y)) -> mark(X) mark(s(X)) -> active(s(mark(X))) active(if(false, X, Y)) -> mark(Y) mark(div(X1, X2)) -> active(div(mark(X1), X2)) active(geq(s(X), s(Y))) -> mark(geq(X, Y)) mark(geq(X1, X2)) -> active(geq(X1, X2)) mark(true) -> active(true) mark(false) -> active(false) div(X1, mark(X2)) -> div(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) active(geq(X, 0)) -> mark(true) active(geq(0, s(Y))) -> mark(false) active(minus(0, Y)) -> mark(0) active(div(0, s(Y))) -> mark(0) if(X1, mark(X2), X3) -> if(X1, X2, X3) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) s(active(X)) -> s(X) s(mark(X)) -> s(X) The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (71) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. MARK(if(X1, X2, X3)) -> ACTIVE(if(mark(X1), X2, X3)) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(ACTIVE(x_1)) = x_1 POL(MARK(x_1)) = 1 + x_1 POL(active(x_1)) = x_1 POL(div(x_1, x_2)) = x_1 + x_2 POL(false) = 1 POL(geq(x_1, x_2)) = 1 POL(if(x_1, x_2, x_3)) = x_1 + x_2 + x_3 POL(mark(x_1)) = x_1 POL(minus(x_1, x_2)) = 0 POL(s(x_1)) = 1 POL(true) = 1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: active(minus(s(X), s(Y))) -> mark(minus(X, Y)) mark(minus(X1, X2)) -> active(minus(X1, X2)) mark(0) -> active(0) active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) mark(if(X1, X2, X3)) -> active(if(mark(X1), X2, X3)) active(if(true, X, Y)) -> mark(X) mark(s(X)) -> active(s(mark(X))) active(if(false, X, Y)) -> mark(Y) mark(div(X1, X2)) -> active(div(mark(X1), X2)) active(geq(s(X), s(Y))) -> mark(geq(X, Y)) mark(geq(X1, X2)) -> active(geq(X1, X2)) mark(true) -> active(true) mark(false) -> active(false) if(X1, mark(X2), X3) -> if(X1, X2, X3) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) active(minus(0, Y)) -> mark(0) active(geq(X, 0)) -> mark(true) active(geq(0, s(Y))) -> mark(false) active(div(0, s(Y))) -> mark(0) s(active(X)) -> s(X) s(mark(X)) -> s(X) div(X1, mark(X2)) -> div(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) ---------------------------------------- (72) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVE(if(true, X, Y)) -> MARK(X) ACTIVE(if(false, X, Y)) -> MARK(Y) MARK(if(X1, X2, X3)) -> MARK(X1) The TRS R consists of the following rules: active(minus(s(X), s(Y))) -> mark(minus(X, Y)) mark(minus(X1, X2)) -> active(minus(X1, X2)) mark(0) -> active(0) active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) mark(if(X1, X2, X3)) -> active(if(mark(X1), X2, X3)) active(if(true, X, Y)) -> mark(X) mark(s(X)) -> active(s(mark(X))) active(if(false, X, Y)) -> mark(Y) mark(div(X1, X2)) -> active(div(mark(X1), X2)) active(geq(s(X), s(Y))) -> mark(geq(X, Y)) mark(geq(X1, X2)) -> active(geq(X1, X2)) mark(true) -> active(true) mark(false) -> active(false) div(X1, mark(X2)) -> div(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) active(geq(X, 0)) -> mark(true) active(geq(0, s(Y))) -> mark(false) active(minus(0, Y)) -> mark(0) active(div(0, s(Y))) -> mark(0) if(X1, mark(X2), X3) -> if(X1, X2, X3) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) s(active(X)) -> s(X) s(mark(X)) -> s(X) The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (73) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. ---------------------------------------- (74) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(if(X1, X2, X3)) -> MARK(X1) The TRS R consists of the following rules: active(minus(s(X), s(Y))) -> mark(minus(X, Y)) mark(minus(X1, X2)) -> active(minus(X1, X2)) mark(0) -> active(0) active(div(s(X), s(Y))) -> mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)) mark(if(X1, X2, X3)) -> active(if(mark(X1), X2, X3)) active(if(true, X, Y)) -> mark(X) mark(s(X)) -> active(s(mark(X))) active(if(false, X, Y)) -> mark(Y) mark(div(X1, X2)) -> active(div(mark(X1), X2)) active(geq(s(X), s(Y))) -> mark(geq(X, Y)) mark(geq(X1, X2)) -> active(geq(X1, X2)) mark(true) -> active(true) mark(false) -> active(false) div(X1, mark(X2)) -> div(X1, X2) div(mark(X1), X2) -> div(X1, X2) div(active(X1), X2) -> div(X1, X2) div(X1, active(X2)) -> div(X1, X2) active(geq(X, 0)) -> mark(true) active(geq(0, s(Y))) -> mark(false) active(minus(0, Y)) -> mark(0) active(div(0, s(Y))) -> mark(0) if(X1, mark(X2), X3) -> if(X1, X2, X3) if(mark(X1), X2, X3) -> if(X1, X2, X3) if(X1, X2, mark(X3)) -> if(X1, X2, X3) if(active(X1), X2, X3) -> if(X1, X2, X3) if(X1, active(X2), X3) -> if(X1, X2, X3) if(X1, X2, active(X3)) -> if(X1, X2, X3) s(active(X)) -> s(X) s(mark(X)) -> s(X) The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (75) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (76) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(if(X1, X2, X3)) -> MARK(X1) R is empty. The set Q consists of the following terms: active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (77) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. active(minus(0, x0)) active(minus(s(x0), s(x1))) active(geq(x0, 0)) active(geq(0, s(x0))) active(geq(s(x0), s(x1))) active(div(0, s(x0))) active(div(s(x0), s(x1))) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(minus(x0, x1)) mark(0) mark(s(x0)) mark(geq(x0, x1)) mark(true) mark(false) mark(div(x0, x1)) mark(if(x0, x1, x2)) minus(mark(x0), x1) minus(x0, mark(x1)) minus(active(x0), x1) minus(x0, active(x1)) s(mark(x0)) s(active(x0)) geq(mark(x0), x1) geq(x0, mark(x1)) geq(active(x0), x1) geq(x0, active(x1)) div(mark(x0), x1) div(x0, mark(x1)) div(active(x0), x1) div(x0, active(x1)) ---------------------------------------- (78) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(if(X1, X2, X3)) -> MARK(X1) R is empty. The set Q consists of the following terms: if(mark(x0), x1, x2) if(x0, mark(x1), x2) if(x0, x1, mark(x2)) if(active(x0), x1, x2) if(x0, active(x1), x2) if(x0, x1, active(x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (79) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *MARK(if(X1, X2, X3)) -> MARK(X1) The graph contains the following edges 1 > 1 ---------------------------------------- (80) YES