YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given ETRS could be proven: (0) ETRS (1) EquationalDependencyPairsProof [EQUIVALENT, 0 ms] (2) EDP (3) EDependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) EDP (6) ESharpUsableEquationsProof [EQUIVALENT, 0 ms] (7) EDP (8) EUsableRulesReductionPairsProof [EQUIVALENT, 8 ms] (9) EDP (10) EDPProblemToQDPProblemProof [EQUIVALENT, 0 ms] (11) QDP (12) MNOCProof [EQUIVALENT, 0 ms] (13) QDP (14) MRRProof [EQUIVALENT, 0 ms] (15) QDP (16) DependencyGraphProof [EQUIVALENT, 0 ms] (17) TRUE (18) EDP (19) ESharpUsableEquationsProof [EQUIVALENT, 0 ms] (20) EDP (21) EDPPoloProof [EQUIVALENT, 0 ms] (22) EDP (23) PisEmptyProof [EQUIVALENT, 0 ms] (24) YES (25) EDP (26) ESharpUsableEquationsProof [EQUIVALENT, 0 ms] (27) EDP (28) EUsableRulesReductionPairsProof [EQUIVALENT, 21 ms] (29) EDP (30) ERuleRemovalProof [EQUIVALENT, 0 ms] (31) EDP (32) EDPPoloProof [EQUIVALENT, 0 ms] (33) EDP (34) PisEmptyProof [EQUIVALENT, 0 ms] (35) YES (36) EDP (37) ESharpUsableEquationsProof [EQUIVALENT, 0 ms] (38) EDP (39) EDPPoloProof [EQUIVALENT, 21 ms] (40) EDP (41) EDPPoloProof [EQUIVALENT, 0 ms] (42) EDP (43) PisEmptyProof [EQUIVALENT, 0 ms] (44) YES ---------------------------------------- (0) Obligation: Equational rewrite system: The TRS R consists of the following rules: p(s(x)) -> x plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) times(x, 0) -> 0 times(x, s(y)) -> plus(x, times(x, y)) minus(x, 0) -> x minus(s(x), s(y)) -> minus(p(s(x)), p(s(y))) div(0, s(y)) -> 0 div(s(x), s(y)) -> s(div(minus(x, y), s(y))) The set E consists of the following equations: plus(x, y) == plus(y, x) times(x, y) == times(y, x) plus(plus(x, y), z) == plus(x, plus(y, z)) times(times(x, y), z) == times(x, times(y, z)) ---------------------------------------- (1) EquationalDependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,DA_STEIN] we result in the following initial EDP problem: The TRS P consists of the following rules: PLUS(x, s(y)) -> PLUS(x, y) TIMES(x, s(y)) -> PLUS(x, times(x, y)) TIMES(x, s(y)) -> TIMES(x, y) MINUS(s(x), s(y)) -> MINUS(p(s(x)), p(s(y))) MINUS(s(x), s(y)) -> P(s(x)) MINUS(s(x), s(y)) -> P(s(y)) DIV(s(x), s(y)) -> DIV(minus(x, y), s(y)) DIV(s(x), s(y)) -> MINUS(x, y) PLUS(plus(x, s(y)), ext) -> PLUS(s(plus(x, y)), ext) PLUS(plus(x, s(y)), ext) -> PLUS(x, y) TIMES(times(x, 0), ext) -> TIMES(0, ext) TIMES(times(x, s(y)), ext) -> TIMES(plus(x, times(x, y)), ext) TIMES(times(x, s(y)), ext) -> PLUS(x, times(x, y)) TIMES(times(x, s(y)), ext) -> TIMES(x, y) The TRS R consists of the following rules: p(s(x)) -> x plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) times(x, 0) -> 0 times(x, s(y)) -> plus(x, times(x, y)) minus(x, 0) -> x minus(s(x), s(y)) -> minus(p(s(x)), p(s(y))) div(0, s(y)) -> 0 div(s(x), s(y)) -> s(div(minus(x, y), s(y))) plus(plus(x, s(y)), ext) -> plus(s(plus(x, y)), ext) times(times(x, 0), ext) -> times(0, ext) times(times(x, s(y)), ext) -> times(plus(x, times(x, y)), ext) The set E consists of the following equations: plus(x, y) == plus(y, x) times(x, y) == times(y, x) plus(plus(x, y), z) == plus(x, plus(y, z)) times(times(x, y), z) == times(x, times(y, z)) The set E# consists of the following equations: PLUS(x, y) == PLUS(y, x) TIMES(x, y) == TIMES(y, x) PLUS(plus(x, y), z) == PLUS(x, plus(y, z)) TIMES(times(x, y), z) == TIMES(x, times(y, z)) We have to consider all minimal (P,E#,R,E)-chains ---------------------------------------- (2) Obligation: The TRS P consists of the following rules: PLUS(x, s(y)) -> PLUS(x, y) TIMES(x, s(y)) -> PLUS(x, times(x, y)) TIMES(x, s(y)) -> TIMES(x, y) MINUS(s(x), s(y)) -> MINUS(p(s(x)), p(s(y))) MINUS(s(x), s(y)) -> P(s(x)) MINUS(s(x), s(y)) -> P(s(y)) DIV(s(x), s(y)) -> DIV(minus(x, y), s(y)) DIV(s(x), s(y)) -> MINUS(x, y) PLUS(plus(x, s(y)), ext) -> PLUS(s(plus(x, y)), ext) PLUS(plus(x, s(y)), ext) -> PLUS(x, y) TIMES(times(x, 0), ext) -> TIMES(0, ext) TIMES(times(x, s(y)), ext) -> TIMES(plus(x, times(x, y)), ext) TIMES(times(x, s(y)), ext) -> PLUS(x, times(x, y)) TIMES(times(x, s(y)), ext) -> TIMES(x, y) The TRS R consists of the following rules: p(s(x)) -> x plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) times(x, 0) -> 0 times(x, s(y)) -> plus(x, times(x, y)) minus(x, 0) -> x minus(s(x), s(y)) -> minus(p(s(x)), p(s(y))) div(0, s(y)) -> 0 div(s(x), s(y)) -> s(div(minus(x, y), s(y))) plus(plus(x, s(y)), ext) -> plus(s(plus(x, y)), ext) times(times(x, 0), ext) -> times(0, ext) times(times(x, s(y)), ext) -> times(plus(x, times(x, y)), ext) The set E consists of the following equations: plus(x, y) == plus(y, x) times(x, y) == times(y, x) plus(plus(x, y), z) == plus(x, plus(y, z)) times(times(x, y), z) == times(x, times(y, z)) The set E# consists of the following equations: PLUS(x, y) == PLUS(y, x) TIMES(x, y) == TIMES(y, x) PLUS(plus(x, y), z) == PLUS(x, plus(y, z)) TIMES(times(x, y), z) == TIMES(x, times(y, z)) We have to consider all minimal (P,E#,R,E)-chains ---------------------------------------- (3) EDependencyGraphProof (EQUIVALENT) The approximation of the Equational Dependency Graph [DA_STEIN] contains 4 SCCs with 5 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: The TRS P consists of the following rules: MINUS(s(x), s(y)) -> MINUS(p(s(x)), p(s(y))) The TRS R consists of the following rules: p(s(x)) -> x plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) times(x, 0) -> 0 times(x, s(y)) -> plus(x, times(x, y)) minus(x, 0) -> x minus(s(x), s(y)) -> minus(p(s(x)), p(s(y))) div(0, s(y)) -> 0 div(s(x), s(y)) -> s(div(minus(x, y), s(y))) plus(plus(x, s(y)), ext) -> plus(s(plus(x, y)), ext) times(times(x, 0), ext) -> times(0, ext) times(times(x, s(y)), ext) -> times(plus(x, times(x, y)), ext) The set E consists of the following equations: plus(x, y) == plus(y, x) times(x, y) == times(y, x) plus(plus(x, y), z) == plus(x, plus(y, z)) times(times(x, y), z) == times(x, times(y, z)) The set E# consists of the following equations: PLUS(x, y) == PLUS(y, x) TIMES(x, y) == TIMES(y, x) PLUS(plus(x, y), z) == PLUS(x, plus(y, z)) TIMES(times(x, y), z) == TIMES(x, times(y, z)) We have to consider all minimal (P,E#,R,E)-chains ---------------------------------------- (6) ESharpUsableEquationsProof (EQUIVALENT) We can delete the following equations of E# with the esharp usable equations processor[DA_STEIN]: PLUS(x, y) == PLUS(y, x) TIMES(x, y) == TIMES(y, x) PLUS(plus(x, y), z) == PLUS(x, plus(y, z)) TIMES(times(x, y), z) == TIMES(x, times(y, z)) ---------------------------------------- (7) Obligation: The TRS P consists of the following rules: MINUS(s(x), s(y)) -> MINUS(p(s(x)), p(s(y))) The TRS R consists of the following rules: p(s(x)) -> x plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) times(x, 0) -> 0 times(x, s(y)) -> plus(x, times(x, y)) minus(x, 0) -> x minus(s(x), s(y)) -> minus(p(s(x)), p(s(y))) div(0, s(y)) -> 0 div(s(x), s(y)) -> s(div(minus(x, y), s(y))) plus(plus(x, s(y)), ext) -> plus(s(plus(x, y)), ext) times(times(x, 0), ext) -> times(0, ext) times(times(x, s(y)), ext) -> times(plus(x, times(x, y)), ext) The set E consists of the following equations: plus(x, y) == plus(y, x) times(x, y) == times(y, x) plus(plus(x, y), z) == plus(x, plus(y, z)) times(times(x, y), z) == times(x, times(y, z)) E# is empty. We have to consider all minimal (P,E#,R,E)-chains ---------------------------------------- (8) EUsableRulesReductionPairsProof (EQUIVALENT) By using the usable rules and equations with reduction pair processor [DA_STEIN] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules can be oriented non-strictly, the usable equations and the esharp equations can be oriented equivalently. All non-usable rules and equations 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. No dependency pairs are removed. The following rules are removed from R: plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) times(x, 0) -> 0 times(x, s(y)) -> plus(x, times(x, y)) minus(x, 0) -> x minus(s(x), s(y)) -> minus(p(s(x)), p(s(y))) div(0, s(y)) -> 0 div(s(x), s(y)) -> s(div(minus(x, y), s(y))) plus(plus(x, s(y)), ext) -> plus(s(plus(x, y)), ext) times(times(x, 0), ext) -> times(0, ext) times(times(x, s(y)), ext) -> times(plus(x, times(x, y)), ext) The following equations are removed from E: plus(x, y) == plus(y, x) times(x, y) == times(y, x) plus(plus(x, y), z) == plus(x, plus(y, z)) times(times(x, y), z) == times(x, times(y, z)) Used ordering: POLO with Polynomial interpretation [POLO]: POL(MINUS(x_1, x_2)) = x_1 + x_2 POL(p(x_1)) = x_1 POL(s(x_1)) = x_1 ---------------------------------------- (9) Obligation: The TRS P consists of the following rules: MINUS(s(x), s(y)) -> MINUS(p(s(x)), p(s(y))) The TRS R consists of the following rules: p(s(x)) -> x E is empty. E# is empty. We have to consider all minimal (P,E#,R,E)-chains ---------------------------------------- (10) EDPProblemToQDPProblemProof (EQUIVALENT) The EDP problem does not contain equations anymore, so we can transform it with the EDP to QDP problem processor [DA_STEIN] into a QDP problem. ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: MINUS(s(x), s(y)) -> MINUS(p(s(x)), p(s(y))) The TRS R consists of the following rules: p(s(x)) -> x Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (12) MNOCProof (EQUIVALENT) We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R. ---------------------------------------- (13) Obligation: Q DP problem: The TRS P consists of the following rules: MINUS(s(x), s(y)) -> MINUS(p(s(x)), p(s(y))) The TRS R consists of the following rules: p(s(x)) -> x The set Q consists of the following terms: p(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (14) 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 rules of the TRS R: p(s(x)) -> x Used ordering: Polynomial interpretation [POLO]: POL(MINUS(x_1, x_2)) = 2*x_1 + 2*x_2 POL(p(x_1)) = x_1 POL(s(x_1)) = 2 + 2*x_1 ---------------------------------------- (15) Obligation: Q DP problem: The TRS P consists of the following rules: MINUS(s(x), s(y)) -> MINUS(p(s(x)), p(s(y))) R is empty. The set Q consists of the following terms: p(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (16) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. ---------------------------------------- (17) TRUE ---------------------------------------- (18) Obligation: The TRS P consists of the following rules: DIV(s(x), s(y)) -> DIV(minus(x, y), s(y)) The TRS R consists of the following rules: p(s(x)) -> x plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) times(x, 0) -> 0 times(x, s(y)) -> plus(x, times(x, y)) minus(x, 0) -> x minus(s(x), s(y)) -> minus(p(s(x)), p(s(y))) div(0, s(y)) -> 0 div(s(x), s(y)) -> s(div(minus(x, y), s(y))) plus(plus(x, s(y)), ext) -> plus(s(plus(x, y)), ext) times(times(x, 0), ext) -> times(0, ext) times(times(x, s(y)), ext) -> times(plus(x, times(x, y)), ext) The set E consists of the following equations: plus(x, y) == plus(y, x) times(x, y) == times(y, x) plus(plus(x, y), z) == plus(x, plus(y, z)) times(times(x, y), z) == times(x, times(y, z)) The set E# consists of the following equations: PLUS(x, y) == PLUS(y, x) TIMES(x, y) == TIMES(y, x) PLUS(plus(x, y), z) == PLUS(x, plus(y, z)) TIMES(times(x, y), z) == TIMES(x, times(y, z)) We have to consider all minimal (P,E#,R,E)-chains ---------------------------------------- (19) ESharpUsableEquationsProof (EQUIVALENT) We can delete the following equations of E# with the esharp usable equations processor[DA_STEIN]: PLUS(x, y) == PLUS(y, x) TIMES(x, y) == TIMES(y, x) PLUS(plus(x, y), z) == PLUS(x, plus(y, z)) TIMES(times(x, y), z) == TIMES(x, times(y, z)) ---------------------------------------- (20) Obligation: The TRS P consists of the following rules: DIV(s(x), s(y)) -> DIV(minus(x, y), s(y)) The TRS R consists of the following rules: p(s(x)) -> x plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) times(x, 0) -> 0 times(x, s(y)) -> plus(x, times(x, y)) minus(x, 0) -> x minus(s(x), s(y)) -> minus(p(s(x)), p(s(y))) div(0, s(y)) -> 0 div(s(x), s(y)) -> s(div(minus(x, y), s(y))) plus(plus(x, s(y)), ext) -> plus(s(plus(x, y)), ext) times(times(x, 0), ext) -> times(0, ext) times(times(x, s(y)), ext) -> times(plus(x, times(x, y)), ext) The set E consists of the following equations: plus(x, y) == plus(y, x) times(x, y) == times(y, x) plus(plus(x, y), z) == plus(x, plus(y, z)) times(times(x, y), z) == times(x, times(y, z)) E# is empty. We have to consider all minimal (P,E#,R,E)-chains ---------------------------------------- (21) EDPPoloProof (EQUIVALENT) We use the reduction pair processor [DA_STEIN] with a polynomial ordering [POLO]. All Dependency Pairs of this DP problem can be strictly oriented. DIV(s(x), s(y)) -> DIV(minus(x, y), s(y)) With the implicit AFS we had to orient the following set of usable rules of R non-strictly. minus(s(x), s(y)) -> minus(p(s(x)), p(s(y))) minus(x, 0) -> x p(s(x)) -> x There is no equation of E#. With the implicit AFS there is no usable equation of E. Used ordering: POLO with Polynomial interpretation [POLO]: POL(0) = 0 POL(DIV(x_1, x_2)) = 2*x_1 POL(minus(x_1, x_2)) = x_1 POL(p(x_1)) = x_1 POL(s(x_1)) = 2 + 2*x_1 ---------------------------------------- (22) Obligation: P is empty. The TRS R consists of the following rules: p(s(x)) -> x plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) times(x, 0) -> 0 times(x, s(y)) -> plus(x, times(x, y)) minus(x, 0) -> x minus(s(x), s(y)) -> minus(p(s(x)), p(s(y))) div(0, s(y)) -> 0 div(s(x), s(y)) -> s(div(minus(x, y), s(y))) plus(plus(x, s(y)), ext) -> plus(s(plus(x, y)), ext) times(times(x, 0), ext) -> times(0, ext) times(times(x, s(y)), ext) -> times(plus(x, times(x, y)), ext) The set E consists of the following equations: plus(x, y) == plus(y, x) times(x, y) == times(y, x) plus(plus(x, y), z) == plus(x, plus(y, z)) times(times(x, y), z) == times(x, times(y, z)) E# is empty. We have to consider all minimal (P,E#,R,E)-chains ---------------------------------------- (23) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,E#,R,E) chain. ---------------------------------------- (24) YES ---------------------------------------- (25) Obligation: The TRS P consists of the following rules: PLUS(plus(x, s(y)), ext) -> PLUS(s(plus(x, y)), ext) PLUS(plus(x, s(y)), ext) -> PLUS(x, y) PLUS(x, s(y)) -> PLUS(x, y) The TRS R consists of the following rules: p(s(x)) -> x plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) times(x, 0) -> 0 times(x, s(y)) -> plus(x, times(x, y)) minus(x, 0) -> x minus(s(x), s(y)) -> minus(p(s(x)), p(s(y))) div(0, s(y)) -> 0 div(s(x), s(y)) -> s(div(minus(x, y), s(y))) plus(plus(x, s(y)), ext) -> plus(s(plus(x, y)), ext) times(times(x, 0), ext) -> times(0, ext) times(times(x, s(y)), ext) -> times(plus(x, times(x, y)), ext) The set E consists of the following equations: plus(x, y) == plus(y, x) times(x, y) == times(y, x) plus(plus(x, y), z) == plus(x, plus(y, z)) times(times(x, y), z) == times(x, times(y, z)) The set E# consists of the following equations: PLUS(x, y) == PLUS(y, x) TIMES(x, y) == TIMES(y, x) PLUS(plus(x, y), z) == PLUS(x, plus(y, z)) TIMES(times(x, y), z) == TIMES(x, times(y, z)) We have to consider all minimal (P,E#,R,E)-chains ---------------------------------------- (26) ESharpUsableEquationsProof (EQUIVALENT) We can delete the following equations of E# with the esharp usable equations processor[DA_STEIN]: TIMES(x, y) == TIMES(y, x) TIMES(times(x, y), z) == TIMES(x, times(y, z)) ---------------------------------------- (27) Obligation: The TRS P consists of the following rules: PLUS(plus(x, s(y)), ext) -> PLUS(s(plus(x, y)), ext) PLUS(plus(x, s(y)), ext) -> PLUS(x, y) PLUS(x, s(y)) -> PLUS(x, y) The TRS R consists of the following rules: p(s(x)) -> x plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) times(x, 0) -> 0 times(x, s(y)) -> plus(x, times(x, y)) minus(x, 0) -> x minus(s(x), s(y)) -> minus(p(s(x)), p(s(y))) div(0, s(y)) -> 0 div(s(x), s(y)) -> s(div(minus(x, y), s(y))) plus(plus(x, s(y)), ext) -> plus(s(plus(x, y)), ext) times(times(x, 0), ext) -> times(0, ext) times(times(x, s(y)), ext) -> times(plus(x, times(x, y)), ext) The set E consists of the following equations: plus(x, y) == plus(y, x) times(x, y) == times(y, x) plus(plus(x, y), z) == plus(x, plus(y, z)) times(times(x, y), z) == times(x, times(y, z)) The set E# consists of the following equations: PLUS(plus(x, y), z) == PLUS(x, plus(y, z)) PLUS(x, y) == PLUS(y, x) We have to consider all minimal (P,E#,R,E)-chains ---------------------------------------- (28) EUsableRulesReductionPairsProof (EQUIVALENT) By using the usable rules and equations with reduction pair processor [DA_STEIN] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules can be oriented non-strictly, the usable equations and the esharp equations can be oriented equivalently. All non-usable rules and equations 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. No dependency pairs are removed. The following rules are removed from R: p(s(x)) -> x plus(x, 0) -> x times(x, 0) -> 0 times(x, s(y)) -> plus(x, times(x, y)) minus(x, 0) -> x minus(s(x), s(y)) -> minus(p(s(x)), p(s(y))) div(0, s(y)) -> 0 div(s(x), s(y)) -> s(div(minus(x, y), s(y))) times(times(x, 0), ext) -> times(0, ext) times(times(x, s(y)), ext) -> times(plus(x, times(x, y)), ext) The following equations are removed from E: times(x, y) == times(y, x) times(times(x, y), z) == times(x, times(y, z)) Used ordering: POLO with Polynomial interpretation [POLO]: POL(0) = 0 POL(PLUS(x_1, x_2)) = x_1 + x_2 POL(plus(x_1, x_2)) = x_1 + x_2 POL(s(x_1)) = x_1 ---------------------------------------- (29) Obligation: The TRS P consists of the following rules: PLUS(plus(x, s(y)), ext) -> PLUS(s(plus(x, y)), ext) PLUS(plus(x, s(y)), ext) -> PLUS(x, y) PLUS(x, s(y)) -> PLUS(x, y) The TRS R consists of the following rules: plus(x, s(y)) -> s(plus(x, y)) plus(plus(x, s(y)), ext) -> plus(s(plus(x, y)), ext) The set E consists of the following equations: plus(plus(x, y), z) == plus(x, plus(y, z)) plus(x, y) == plus(y, x) The set E# consists of the following equations: PLUS(plus(x, y), z) == PLUS(x, plus(y, z)) PLUS(x, y) == PLUS(y, x) We have to consider all minimal (P,E#,R,E)-chains ---------------------------------------- (30) ERuleRemovalProof (EQUIVALENT) By using the rule removal processor [DA_STEIN] with the following polynomial ordering [POLO], at least one Dependency Pair or term rewrite system rule of this EDP problem can be strictly oriented. Strictly oriented dependency pairs: PLUS(plus(x, s(y)), ext) -> PLUS(x, y) PLUS(x, s(y)) -> PLUS(x, y) Used ordering: POLO with Polynomial interpretation [POLO]: POL(PLUS(x_1, x_2)) = x_1 + x_2 POL(plus(x_1, x_2)) = 2 + x_1 + x_2 POL(s(x_1)) = 2 + x_1 ---------------------------------------- (31) Obligation: The TRS P consists of the following rules: PLUS(plus(x, s(y)), ext) -> PLUS(s(plus(x, y)), ext) The TRS R consists of the following rules: plus(x, s(y)) -> s(plus(x, y)) plus(plus(x, s(y)), ext) -> plus(s(plus(x, y)), ext) The set E consists of the following equations: plus(plus(x, y), z) == plus(x, plus(y, z)) plus(x, y) == plus(y, x) The set E# consists of the following equations: PLUS(plus(x, y), z) == PLUS(x, plus(y, z)) PLUS(x, y) == PLUS(y, x) We have to consider all minimal (P,E#,R,E)-chains ---------------------------------------- (32) EDPPoloProof (EQUIVALENT) We use the reduction pair processor [DA_STEIN] with a polynomial ordering [POLO]. All Dependency Pairs of this DP problem can be strictly oriented. PLUS(plus(x, s(y)), ext) -> PLUS(s(plus(x, y)), ext) With the implicit AFS we had to orient the following set of usable rules of R non-strictly. plus(x, s(y)) -> s(plus(x, y)) plus(plus(x, s(y)), ext) -> plus(s(plus(x, y)), ext) We had to orient the following equations of E# equivalently. PLUS(plus(x, y), z) == PLUS(x, plus(y, z)) PLUS(x, y) == PLUS(y, x) With the implicit AFS we had to orient the following usable equations of E equivalently. plus(plus(x, y), z) == plus(x, plus(y, z)) plus(x, y) == plus(y, x) Used ordering: POLO with Polynomial interpretation [POLO]: POL(PLUS(x_1, x_2)) = x_1 + x_2 POL(plus(x_1, x_2)) = 1 + x_1 + x_2 POL(s(x_1)) = 0 ---------------------------------------- (33) Obligation: P is empty. The TRS R consists of the following rules: plus(x, s(y)) -> s(plus(x, y)) plus(plus(x, s(y)), ext) -> plus(s(plus(x, y)), ext) The set E consists of the following equations: plus(plus(x, y), z) == plus(x, plus(y, z)) plus(x, y) == plus(y, x) The set E# consists of the following equations: PLUS(plus(x, y), z) == PLUS(x, plus(y, z)) PLUS(x, y) == PLUS(y, x) We have to consider all minimal (P,E#,R,E)-chains ---------------------------------------- (34) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,E#,R,E) chain. ---------------------------------------- (35) YES ---------------------------------------- (36) Obligation: The TRS P consists of the following rules: TIMES(times(x, s(y)), ext) -> TIMES(plus(x, times(x, y)), ext) TIMES(times(x, s(y)), ext) -> TIMES(x, y) TIMES(x, s(y)) -> TIMES(x, y) TIMES(times(x, 0), ext) -> TIMES(0, ext) The TRS R consists of the following rules: p(s(x)) -> x plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) times(x, 0) -> 0 times(x, s(y)) -> plus(x, times(x, y)) minus(x, 0) -> x minus(s(x), s(y)) -> minus(p(s(x)), p(s(y))) div(0, s(y)) -> 0 div(s(x), s(y)) -> s(div(minus(x, y), s(y))) plus(plus(x, s(y)), ext) -> plus(s(plus(x, y)), ext) times(times(x, 0), ext) -> times(0, ext) times(times(x, s(y)), ext) -> times(plus(x, times(x, y)), ext) The set E consists of the following equations: plus(x, y) == plus(y, x) times(x, y) == times(y, x) plus(plus(x, y), z) == plus(x, plus(y, z)) times(times(x, y), z) == times(x, times(y, z)) The set E# consists of the following equations: PLUS(x, y) == PLUS(y, x) TIMES(x, y) == TIMES(y, x) PLUS(plus(x, y), z) == PLUS(x, plus(y, z)) TIMES(times(x, y), z) == TIMES(x, times(y, z)) We have to consider all minimal (P,E#,R,E)-chains ---------------------------------------- (37) ESharpUsableEquationsProof (EQUIVALENT) We can delete the following equations of E# with the esharp usable equations processor[DA_STEIN]: PLUS(x, y) == PLUS(y, x) PLUS(plus(x, y), z) == PLUS(x, plus(y, z)) ---------------------------------------- (38) Obligation: The TRS P consists of the following rules: TIMES(times(x, s(y)), ext) -> TIMES(plus(x, times(x, y)), ext) TIMES(times(x, s(y)), ext) -> TIMES(x, y) TIMES(x, s(y)) -> TIMES(x, y) TIMES(times(x, 0), ext) -> TIMES(0, ext) The TRS R consists of the following rules: p(s(x)) -> x plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) times(x, 0) -> 0 times(x, s(y)) -> plus(x, times(x, y)) minus(x, 0) -> x minus(s(x), s(y)) -> minus(p(s(x)), p(s(y))) div(0, s(y)) -> 0 div(s(x), s(y)) -> s(div(minus(x, y), s(y))) plus(plus(x, s(y)), ext) -> plus(s(plus(x, y)), ext) times(times(x, 0), ext) -> times(0, ext) times(times(x, s(y)), ext) -> times(plus(x, times(x, y)), ext) The set E consists of the following equations: plus(x, y) == plus(y, x) times(x, y) == times(y, x) plus(plus(x, y), z) == plus(x, plus(y, z)) times(times(x, y), z) == times(x, times(y, z)) The set E# consists of the following equations: TIMES(times(x, y), z) == TIMES(x, times(y, z)) TIMES(x, y) == TIMES(y, x) We have to consider all minimal (P,E#,R,E)-chains ---------------------------------------- (39) EDPPoloProof (EQUIVALENT) We use the reduction pair processor [DA_STEIN] with a polynomial ordering [POLO]. The following set of Dependency Pairs of this DP problem can be strictly oriented. TIMES(times(x, s(y)), ext) -> TIMES(plus(x, times(x, y)), ext) TIMES(times(x, s(y)), ext) -> TIMES(x, y) TIMES(x, s(y)) -> TIMES(x, y) The remaining Dependency Pairs were at least non-strictly oriented. TIMES(times(x, 0), ext) -> TIMES(0, ext) With the implicit AFS we had to orient the following set of usable rules of R non-strictly. times(x, s(y)) -> plus(x, times(x, y)) times(times(x, s(y)), ext) -> times(plus(x, times(x, y)), ext) times(times(x, 0), ext) -> times(0, ext) times(x, 0) -> 0 plus(x, s(y)) -> s(plus(x, y)) plus(plus(x, s(y)), ext) -> plus(s(plus(x, y)), ext) plus(x, 0) -> x We had to orient the following equations of E# equivalently. TIMES(times(x, y), z) == TIMES(x, times(y, z)) TIMES(x, y) == TIMES(y, x) With the implicit AFS we had to orient the following usable equations of E equivalently. times(x, y) == times(y, x) times(times(x, y), z) == times(x, times(y, z)) plus(plus(x, y), z) == plus(x, plus(y, z)) plus(x, y) == plus(y, x) Used ordering: POLO with Polynomial interpretation [POLO]: POL(0) = 0 POL(TIMES(x_1, x_2)) = x_1 + x_1*x_2 + x_2 POL(plus(x_1, x_2)) = x_1 + x_2 POL(s(x_1)) = 1 + x_1 POL(times(x_1, x_2)) = x_1 + x_1*x_2 + x_2 ---------------------------------------- (40) Obligation: The TRS P consists of the following rules: TIMES(times(x, 0), ext) -> TIMES(0, ext) The TRS R consists of the following rules: p(s(x)) -> x plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) times(x, 0) -> 0 times(x, s(y)) -> plus(x, times(x, y)) minus(x, 0) -> x minus(s(x), s(y)) -> minus(p(s(x)), p(s(y))) div(0, s(y)) -> 0 div(s(x), s(y)) -> s(div(minus(x, y), s(y))) plus(plus(x, s(y)), ext) -> plus(s(plus(x, y)), ext) times(times(x, 0), ext) -> times(0, ext) times(times(x, s(y)), ext) -> times(plus(x, times(x, y)), ext) The set E consists of the following equations: plus(x, y) == plus(y, x) times(x, y) == times(y, x) plus(plus(x, y), z) == plus(x, plus(y, z)) times(times(x, y), z) == times(x, times(y, z)) The set E# consists of the following equations: TIMES(times(x, y), z) == TIMES(x, times(y, z)) TIMES(x, y) == TIMES(y, x) We have to consider all minimal (P,E#,R,E)-chains ---------------------------------------- (41) EDPPoloProof (EQUIVALENT) We use the reduction pair processor [DA_STEIN] with a polynomial ordering [POLO]. All Dependency Pairs of this DP problem can be strictly oriented. TIMES(times(x, 0), ext) -> TIMES(0, ext) With the implicit AFS we had to orient the following set of usable rules of R non-strictly. times(x, s(y)) -> plus(x, times(x, y)) times(times(x, s(y)), ext) -> times(plus(x, times(x, y)), ext) times(times(x, 0), ext) -> times(0, ext) times(x, 0) -> 0 plus(x, s(y)) -> s(plus(x, y)) plus(plus(x, s(y)), ext) -> plus(s(plus(x, y)), ext) plus(x, 0) -> x We had to orient the following equations of E# equivalently. TIMES(times(x, y), z) == TIMES(x, times(y, z)) TIMES(x, y) == TIMES(y, x) With the implicit AFS we had to orient the following usable equations of E equivalently. times(x, y) == times(y, x) times(times(x, y), z) == times(x, times(y, z)) plus(plus(x, y), z) == plus(x, plus(y, z)) plus(x, y) == plus(y, x) Used ordering: POLO with Polynomial interpretation [POLO]: POL(0) = 2 POL(TIMES(x_1, x_2)) = 2*x_1 + x_1*x_2 + 2*x_2 POL(plus(x_1, x_2)) = x_1 + x_2 POL(s(x_1)) = 2 + x_1 POL(times(x_1, x_2)) = 2 + 2*x_1 + x_1*x_2 + 2*x_2 ---------------------------------------- (42) Obligation: P is empty. The TRS R consists of the following rules: p(s(x)) -> x plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) times(x, 0) -> 0 times(x, s(y)) -> plus(x, times(x, y)) minus(x, 0) -> x minus(s(x), s(y)) -> minus(p(s(x)), p(s(y))) div(0, s(y)) -> 0 div(s(x), s(y)) -> s(div(minus(x, y), s(y))) plus(plus(x, s(y)), ext) -> plus(s(plus(x, y)), ext) times(times(x, 0), ext) -> times(0, ext) times(times(x, s(y)), ext) -> times(plus(x, times(x, y)), ext) The set E consists of the following equations: plus(x, y) == plus(y, x) times(x, y) == times(y, x) plus(plus(x, y), z) == plus(x, plus(y, z)) times(times(x, y), z) == times(x, times(y, z)) The set E# consists of the following equations: TIMES(times(x, y), z) == TIMES(x, times(y, z)) TIMES(x, y) == TIMES(y, x) We have to consider all minimal (P,E#,R,E)-chains ---------------------------------------- (43) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,E#,R,E) chain. ---------------------------------------- (44) YES