YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 86 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 0 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 18 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) AND (9) QDP (10) UsableRulesProof [EQUIVALENT, 0 ms] (11) QDP (12) QReductionProof [EQUIVALENT, 0 ms] (13) QDP (14) QDPSizeChangeProof [EQUIVALENT, 0 ms] (15) YES (16) QDP (17) UsableRulesProof [EQUIVALENT, 0 ms] (18) QDP (19) QReductionProof [EQUIVALENT, 0 ms] (20) QDP (21) QDPSizeChangeProof [EQUIVALENT, 0 ms] (22) YES (23) QDP (24) MRRProof [EQUIVALENT, 21 ms] (25) QDP (26) QDPOrderProof [EQUIVALENT, 17 ms] (27) QDP (28) DependencyGraphProof [EQUIVALENT, 0 ms] (29) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: active(f(X)) -> mark(if(X, c, f(true))) active(if(true, X, Y)) -> mark(X) active(if(false, X, Y)) -> mark(Y) mark(f(X)) -> active(f(mark(X))) mark(if(X1, X2, X3)) -> active(if(mark(X1), mark(X2), X3)) mark(c) -> active(c) mark(true) -> active(true) mark(false) -> active(false) f(mark(X)) -> f(X) f(active(X)) -> f(X) 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(f(x0)) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(f(x0)) mark(if(x0, x1, x2)) mark(c) mark(true) mark(false) f(mark(x0)) f(active(x0)) 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) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(active(x_1)) = x_1 POL(c) = 0 POL(f(x_1)) = x_1 POL(false) = 1 POL(if(x_1, x_2, x_3)) = x_1 + x_2 + x_3 POL(mark(x_1)) = x_1 POL(true) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: active(if(false, X, Y)) -> mark(Y) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: active(f(X)) -> mark(if(X, c, f(true))) active(if(true, X, Y)) -> mark(X) mark(f(X)) -> active(f(mark(X))) mark(if(X1, X2, X3)) -> active(if(mark(X1), mark(X2), X3)) mark(c) -> active(c) mark(true) -> active(true) mark(false) -> active(false) f(mark(X)) -> f(X) f(active(X)) -> f(X) 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(f(x0)) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(f(x0)) mark(if(x0, x1, x2)) mark(c) mark(true) mark(false) f(mark(x0)) f(active(x0)) 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)) ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(active(x_1)) = x_1 POL(c) = 0 POL(f(x_1)) = 2*x_1 POL(false) = 1 POL(if(x_1, x_2, x_3)) = x_1 + 2*x_2 + 2*x_3 POL(mark(x_1)) = 2*x_1 POL(true) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: mark(false) -> active(false) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: active(f(X)) -> mark(if(X, c, f(true))) active(if(true, X, Y)) -> mark(X) mark(f(X)) -> active(f(mark(X))) mark(if(X1, X2, X3)) -> active(if(mark(X1), mark(X2), X3)) mark(c) -> active(c) mark(true) -> active(true) f(mark(X)) -> f(X) f(active(X)) -> f(X) 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(f(x0)) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(f(x0)) mark(if(x0, x1, x2)) mark(c) mark(true) mark(false) f(mark(x0)) f(active(x0)) 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)) ---------------------------------------- (5) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVE(f(X)) -> MARK(if(X, c, f(true))) ACTIVE(f(X)) -> IF(X, c, f(true)) ACTIVE(f(X)) -> F(true) ACTIVE(if(true, X, Y)) -> MARK(X) MARK(f(X)) -> ACTIVE(f(mark(X))) MARK(f(X)) -> F(mark(X)) MARK(f(X)) -> MARK(X) MARK(if(X1, X2, X3)) -> ACTIVE(if(mark(X1), mark(X2), X3)) MARK(if(X1, X2, X3)) -> IF(mark(X1), mark(X2), X3) MARK(if(X1, X2, X3)) -> MARK(X1) MARK(if(X1, X2, X3)) -> MARK(X2) MARK(c) -> ACTIVE(c) MARK(true) -> ACTIVE(true) F(mark(X)) -> F(X) F(active(X)) -> F(X) 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(f(X)) -> mark(if(X, c, f(true))) active(if(true, X, Y)) -> mark(X) mark(f(X)) -> active(f(mark(X))) mark(if(X1, X2, X3)) -> active(if(mark(X1), mark(X2), X3)) mark(c) -> active(c) mark(true) -> active(true) f(mark(X)) -> f(X) f(active(X)) -> f(X) 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(f(x0)) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(f(x0)) mark(if(x0, x1, x2)) mark(c) mark(true) mark(false) f(mark(x0)) f(active(x0)) 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. ---------------------------------------- (7) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 6 less nodes. ---------------------------------------- (8) Complex Obligation (AND) ---------------------------------------- (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) The TRS R consists of the following rules: active(f(X)) -> mark(if(X, c, f(true))) active(if(true, X, Y)) -> mark(X) mark(f(X)) -> active(f(mark(X))) mark(if(X1, X2, X3)) -> active(if(mark(X1), mark(X2), X3)) mark(c) -> active(c) mark(true) -> active(true) f(mark(X)) -> f(X) f(active(X)) -> f(X) 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(f(x0)) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(f(x0)) mark(if(x0, x1, x2)) mark(c) mark(true) mark(false) f(mark(x0)) f(active(x0)) 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. ---------------------------------------- (10) 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. ---------------------------------------- (11) 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(f(x0)) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(f(x0)) mark(if(x0, x1, x2)) mark(c) mark(true) mark(false) f(mark(x0)) f(active(x0)) 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. ---------------------------------------- (12) 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]. f(mark(x0)) f(active(x0)) 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)) ---------------------------------------- (13) 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(f(x0)) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(f(x0)) mark(if(x0, x1, x2)) mark(c) mark(true) mark(false) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (14) 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 ---------------------------------------- (15) YES ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: F(active(X)) -> F(X) F(mark(X)) -> F(X) The TRS R consists of the following rules: active(f(X)) -> mark(if(X, c, f(true))) active(if(true, X, Y)) -> mark(X) mark(f(X)) -> active(f(mark(X))) mark(if(X1, X2, X3)) -> active(if(mark(X1), mark(X2), X3)) mark(c) -> active(c) mark(true) -> active(true) f(mark(X)) -> f(X) f(active(X)) -> f(X) 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(f(x0)) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(f(x0)) mark(if(x0, x1, x2)) mark(c) mark(true) mark(false) f(mark(x0)) f(active(x0)) 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. ---------------------------------------- (17) 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. ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: F(active(X)) -> F(X) F(mark(X)) -> F(X) R is empty. The set Q consists of the following terms: active(f(x0)) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(f(x0)) mark(if(x0, x1, x2)) mark(c) mark(true) mark(false) f(mark(x0)) f(active(x0)) 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. ---------------------------------------- (19) 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]. f(mark(x0)) f(active(x0)) 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)) ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: F(active(X)) -> F(X) F(mark(X)) -> F(X) R is empty. The set Q consists of the following terms: active(f(x0)) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(f(x0)) mark(if(x0, x1, x2)) mark(c) mark(true) mark(false) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) 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: *F(active(X)) -> F(X) The graph contains the following edges 1 > 1 *F(mark(X)) -> F(X) The graph contains the following edges 1 > 1 ---------------------------------------- (22) YES ---------------------------------------- (23) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(if(X1, X2, X3)) -> ACTIVE(if(mark(X1), mark(X2), X3)) ACTIVE(f(X)) -> MARK(if(X, c, f(true))) MARK(if(X1, X2, X3)) -> MARK(X1) MARK(f(X)) -> ACTIVE(f(mark(X))) ACTIVE(if(true, X, Y)) -> MARK(X) MARK(f(X)) -> MARK(X) MARK(if(X1, X2, X3)) -> MARK(X2) The TRS R consists of the following rules: active(f(X)) -> mark(if(X, c, f(true))) active(if(true, X, Y)) -> mark(X) mark(f(X)) -> active(f(mark(X))) mark(if(X1, X2, X3)) -> active(if(mark(X1), mark(X2), X3)) mark(c) -> active(c) mark(true) -> active(true) f(mark(X)) -> f(X) f(active(X)) -> f(X) 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(f(x0)) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(f(x0)) mark(if(x0, x1, x2)) mark(c) mark(true) mark(false) f(mark(x0)) f(active(x0)) 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. ---------------------------------------- (24) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: MARK(f(X)) -> MARK(X) Used ordering: Polynomial interpretation [POLO]: POL(ACTIVE(x_1)) = 2 + x_1 POL(MARK(x_1)) = 2 + x_1 POL(active(x_1)) = x_1 POL(c) = 0 POL(f(x_1)) = 2 + 2*x_1 POL(if(x_1, x_2, x_3)) = 2*x_1 + x_2 + x_3 POL(mark(x_1)) = x_1 POL(true) = 0 ---------------------------------------- (25) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(if(X1, X2, X3)) -> ACTIVE(if(mark(X1), mark(X2), X3)) ACTIVE(f(X)) -> MARK(if(X, c, f(true))) MARK(if(X1, X2, X3)) -> MARK(X1) MARK(f(X)) -> ACTIVE(f(mark(X))) ACTIVE(if(true, X, Y)) -> MARK(X) MARK(if(X1, X2, X3)) -> MARK(X2) The TRS R consists of the following rules: active(f(X)) -> mark(if(X, c, f(true))) active(if(true, X, Y)) -> mark(X) mark(f(X)) -> active(f(mark(X))) mark(if(X1, X2, X3)) -> active(if(mark(X1), mark(X2), X3)) mark(c) -> active(c) mark(true) -> active(true) f(mark(X)) -> f(X) f(active(X)) -> f(X) 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(f(x0)) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(f(x0)) mark(if(x0, x1, x2)) mark(c) mark(true) mark(false) f(mark(x0)) f(active(x0)) 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. ---------------------------------------- (26) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. ACTIVE(f(X)) -> MARK(if(X, c, f(true))) MARK(if(X1, X2, X3)) -> MARK(X1) ACTIVE(if(true, X, Y)) -> MARK(X) MARK(if(X1, X2, X3)) -> MARK(X2) The remaining pairs can at least be oriented weakly. Used ordering: Combined order from the following AFS and order. MARK(x1) = x1 if(x1, x2, x3) = if(x1, x2) ACTIVE(x1) = x1 mark(x1) = x1 f(x1) = f(x1) c = c true = true active(x1) = x1 Knuth-Bendix order [KBO] with precedence:trivial and weight map: true=1 c=2 f_1=4 if_2=1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: mark(if(X1, X2, X3)) -> active(if(mark(X1), mark(X2), X3)) active(f(X)) -> mark(if(X, c, f(true))) active(if(true, X, Y)) -> mark(X) mark(f(X)) -> active(f(mark(X))) mark(c) -> active(c) mark(true) -> active(true) 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) f(active(X)) -> f(X) f(mark(X)) -> f(X) ---------------------------------------- (27) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(if(X1, X2, X3)) -> ACTIVE(if(mark(X1), mark(X2), X3)) MARK(f(X)) -> ACTIVE(f(mark(X))) The TRS R consists of the following rules: active(f(X)) -> mark(if(X, c, f(true))) active(if(true, X, Y)) -> mark(X) mark(f(X)) -> active(f(mark(X))) mark(if(X1, X2, X3)) -> active(if(mark(X1), mark(X2), X3)) mark(c) -> active(c) mark(true) -> active(true) f(mark(X)) -> f(X) f(active(X)) -> f(X) 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(f(x0)) active(if(true, x0, x1)) active(if(false, x0, x1)) mark(f(x0)) mark(if(x0, x1, x2)) mark(c) mark(true) mark(false) f(mark(x0)) f(active(x0)) 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. ---------------------------------------- (28) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes. ---------------------------------------- (29) TRUE