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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 11 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) MNOCProof [EQUIVALENT, 8 ms] (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPSizeChangeProof [EQUIVALENT, 0 ms] (18) YES (19) QDP (20) MNOCProof [EQUIVALENT, 0 ms] (21) QDP (22) UsableRulesProof [EQUIVALENT, 0 ms] (23) QDP (24) QReductionProof [EQUIVALENT, 0 ms] (25) QDP (26) QDPOrderProof [EQUIVALENT, 9 ms] (27) QDP (28) QDPOrderProof [EQUIVALENT, 1 ms] (29) QDP (30) PisEmptyProof [EQUIVALENT, 0 ms] (31) YES (32) QDP (33) MNOCProof [EQUIVALENT, 0 ms] (34) QDP (35) UsableRulesProof [EQUIVALENT, 0 ms] (36) QDP (37) QReductionProof [EQUIVALENT, 0 ms] (38) QDP (39) QDPOrderProof [EQUIVALENT, 58 ms] (40) QDP (41) DependencyGraphProof [EQUIVALENT, 0 ms] (42) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: p(0(x1)) -> 0(s(s(p(x1)))) p(s(x1)) -> x1 p(p(s(x1))) -> p(x1) f(s(x1)) -> p(s(g(p(s(s(x1)))))) g(s(x1)) -> p(p(s(s(s(j(s(p(s(p(s(x1))))))))))) j(s(x1)) -> p(s(s(p(s(f(p(s(p(p(s(x1))))))))))) half(0(x1)) -> 0(s(s(half(p(s(p(s(x1)))))))) half(s(s(x1))) -> s(half(p(p(s(s(x1)))))) rd(0(x1)) -> 0(s(0(0(0(0(s(0(rd(x1))))))))) Q is empty. ---------------------------------------- (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: P(0(x1)) -> P(x1) P(p(s(x1))) -> P(x1) F(s(x1)) -> P(s(g(p(s(s(x1)))))) F(s(x1)) -> G(p(s(s(x1)))) F(s(x1)) -> P(s(s(x1))) G(s(x1)) -> P(p(s(s(s(j(s(p(s(p(s(x1))))))))))) G(s(x1)) -> P(s(s(s(j(s(p(s(p(s(x1)))))))))) G(s(x1)) -> J(s(p(s(p(s(x1)))))) G(s(x1)) -> P(s(p(s(x1)))) G(s(x1)) -> P(s(x1)) J(s(x1)) -> P(s(s(p(s(f(p(s(p(p(s(x1))))))))))) J(s(x1)) -> P(s(f(p(s(p(p(s(x1)))))))) J(s(x1)) -> F(p(s(p(p(s(x1)))))) J(s(x1)) -> P(s(p(p(s(x1))))) J(s(x1)) -> P(p(s(x1))) J(s(x1)) -> P(s(x1)) HALF(0(x1)) -> HALF(p(s(p(s(x1))))) HALF(0(x1)) -> P(s(p(s(x1)))) HALF(0(x1)) -> P(s(x1)) HALF(s(s(x1))) -> HALF(p(p(s(s(x1))))) HALF(s(s(x1))) -> P(p(s(s(x1)))) HALF(s(s(x1))) -> P(s(s(x1))) RD(0(x1)) -> RD(x1) The TRS R consists of the following rules: p(0(x1)) -> 0(s(s(p(x1)))) p(s(x1)) -> x1 p(p(s(x1))) -> p(x1) f(s(x1)) -> p(s(g(p(s(s(x1)))))) g(s(x1)) -> p(p(s(s(s(j(s(p(s(p(s(x1))))))))))) j(s(x1)) -> p(s(s(p(s(f(p(s(p(p(s(x1))))))))))) half(0(x1)) -> 0(s(s(half(p(s(p(s(x1)))))))) half(s(s(x1))) -> s(half(p(p(s(s(x1)))))) rd(0(x1)) -> 0(s(0(0(0(0(s(0(rd(x1))))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 15 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: RD(0(x1)) -> RD(x1) The TRS R consists of the following rules: p(0(x1)) -> 0(s(s(p(x1)))) p(s(x1)) -> x1 p(p(s(x1))) -> p(x1) f(s(x1)) -> p(s(g(p(s(s(x1)))))) g(s(x1)) -> p(p(s(s(s(j(s(p(s(p(s(x1))))))))))) j(s(x1)) -> p(s(s(p(s(f(p(s(p(p(s(x1))))))))))) half(0(x1)) -> 0(s(s(half(p(s(p(s(x1)))))))) half(s(s(x1))) -> s(half(p(p(s(s(x1)))))) rd(0(x1)) -> 0(s(0(0(0(0(s(0(rd(x1))))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (6) MNOCProof (EQUIVALENT) We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R. ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: RD(0(x1)) -> RD(x1) The TRS R consists of the following rules: p(0(x1)) -> 0(s(s(p(x1)))) p(s(x1)) -> x1 p(p(s(x1))) -> p(x1) f(s(x1)) -> p(s(g(p(s(s(x1)))))) g(s(x1)) -> p(p(s(s(s(j(s(p(s(p(s(x1))))))))))) j(s(x1)) -> p(s(s(p(s(f(p(s(p(p(s(x1))))))))))) half(0(x1)) -> 0(s(s(half(p(s(p(s(x1)))))))) half(s(s(x1))) -> s(half(p(p(s(s(x1)))))) rd(0(x1)) -> 0(s(0(0(0(0(s(0(rd(x1))))))))) The set Q consists of the following terms: p(0(x0)) p(s(x0)) f(s(x0)) g(s(x0)) j(s(x0)) half(0(x0)) half(s(s(x0))) rd(0(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) 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. ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: RD(0(x1)) -> RD(x1) R is empty. The set Q consists of the following terms: p(0(x0)) p(s(x0)) f(s(x0)) g(s(x0)) j(s(x0)) half(0(x0)) half(s(s(x0))) rd(0(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) 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]. p(0(x0)) p(s(x0)) f(s(x0)) g(s(x0)) j(s(x0)) half(0(x0)) half(s(s(x0))) rd(0(x0)) ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: RD(0(x1)) -> RD(x1) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (12) 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: *RD(0(x1)) -> RD(x1) The graph contains the following edges 1 > 1 ---------------------------------------- (13) YES ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: P(p(s(x1))) -> P(x1) P(0(x1)) -> P(x1) The TRS R consists of the following rules: p(0(x1)) -> 0(s(s(p(x1)))) p(s(x1)) -> x1 p(p(s(x1))) -> p(x1) f(s(x1)) -> p(s(g(p(s(s(x1)))))) g(s(x1)) -> p(p(s(s(s(j(s(p(s(p(s(x1))))))))))) j(s(x1)) -> p(s(s(p(s(f(p(s(p(p(s(x1))))))))))) half(0(x1)) -> 0(s(s(half(p(s(p(s(x1)))))))) half(s(s(x1))) -> s(half(p(p(s(s(x1)))))) rd(0(x1)) -> 0(s(0(0(0(0(s(0(rd(x1))))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: P(p(s(x1))) -> P(x1) P(0(x1)) -> P(x1) R is empty. Q is empty. 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: *P(p(s(x1))) -> P(x1) The graph contains the following edges 1 > 1 *P(0(x1)) -> P(x1) The graph contains the following edges 1 > 1 ---------------------------------------- (18) YES ---------------------------------------- (19) Obligation: Q DP problem: The TRS P consists of the following rules: HALF(s(s(x1))) -> HALF(p(p(s(s(x1))))) HALF(0(x1)) -> HALF(p(s(p(s(x1))))) The TRS R consists of the following rules: p(0(x1)) -> 0(s(s(p(x1)))) p(s(x1)) -> x1 p(p(s(x1))) -> p(x1) f(s(x1)) -> p(s(g(p(s(s(x1)))))) g(s(x1)) -> p(p(s(s(s(j(s(p(s(p(s(x1))))))))))) j(s(x1)) -> p(s(s(p(s(f(p(s(p(p(s(x1))))))))))) half(0(x1)) -> 0(s(s(half(p(s(p(s(x1)))))))) half(s(s(x1))) -> s(half(p(p(s(s(x1)))))) rd(0(x1)) -> 0(s(0(0(0(0(s(0(rd(x1))))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (20) MNOCProof (EQUIVALENT) We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R. ---------------------------------------- (21) Obligation: Q DP problem: The TRS P consists of the following rules: HALF(s(s(x1))) -> HALF(p(p(s(s(x1))))) HALF(0(x1)) -> HALF(p(s(p(s(x1))))) The TRS R consists of the following rules: p(0(x1)) -> 0(s(s(p(x1)))) p(s(x1)) -> x1 p(p(s(x1))) -> p(x1) f(s(x1)) -> p(s(g(p(s(s(x1)))))) g(s(x1)) -> p(p(s(s(s(j(s(p(s(p(s(x1))))))))))) j(s(x1)) -> p(s(s(p(s(f(p(s(p(p(s(x1))))))))))) half(0(x1)) -> 0(s(s(half(p(s(p(s(x1)))))))) half(s(s(x1))) -> s(half(p(p(s(s(x1)))))) rd(0(x1)) -> 0(s(0(0(0(0(s(0(rd(x1))))))))) The set Q consists of the following terms: p(0(x0)) p(s(x0)) f(s(x0)) g(s(x0)) j(s(x0)) half(0(x0)) half(s(s(x0))) rd(0(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (22) 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. ---------------------------------------- (23) Obligation: Q DP problem: The TRS P consists of the following rules: HALF(s(s(x1))) -> HALF(p(p(s(s(x1))))) HALF(0(x1)) -> HALF(p(s(p(s(x1))))) The TRS R consists of the following rules: p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(p(x1)))) The set Q consists of the following terms: p(0(x0)) p(s(x0)) f(s(x0)) g(s(x0)) j(s(x0)) half(0(x0)) half(s(s(x0))) rd(0(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (24) 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(s(x0)) g(s(x0)) j(s(x0)) half(0(x0)) half(s(s(x0))) rd(0(x0)) ---------------------------------------- (25) Obligation: Q DP problem: The TRS P consists of the following rules: HALF(s(s(x1))) -> HALF(p(p(s(s(x1))))) HALF(0(x1)) -> HALF(p(s(p(s(x1))))) The TRS R consists of the following rules: p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(p(x1)))) The set Q consists of the following terms: p(0(x0)) p(s(x0)) 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. HALF(0(x1)) -> HALF(p(s(p(s(x1))))) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = 1 + x_1 POL(HALF(x_1)) = x_1 POL(p(x_1)) = x_1 POL(s(x_1)) = x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(p(x1)))) ---------------------------------------- (27) Obligation: Q DP problem: The TRS P consists of the following rules: HALF(s(s(x1))) -> HALF(p(p(s(s(x1))))) The TRS R consists of the following rules: p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(p(x1)))) The set Q consists of the following terms: p(0(x0)) p(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (28) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. HALF(s(s(x1))) -> HALF(p(p(s(s(x1))))) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( HALF_1(x_1) ) = 2x_1 + 2 POL( p_1(x_1) ) = max{0, x_1 - 2} POL( s_1(x_1) ) = 2x_1 + 2 POL( 0_1(x_1) ) = max{0, -2} The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(p(x1)))) ---------------------------------------- (29) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(p(x1)))) The set Q consists of the following terms: p(0(x0)) p(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (30) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (31) YES ---------------------------------------- (32) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(x1)) -> G(p(s(s(x1)))) G(s(x1)) -> J(s(p(s(p(s(x1)))))) J(s(x1)) -> F(p(s(p(p(s(x1)))))) The TRS R consists of the following rules: p(0(x1)) -> 0(s(s(p(x1)))) p(s(x1)) -> x1 p(p(s(x1))) -> p(x1) f(s(x1)) -> p(s(g(p(s(s(x1)))))) g(s(x1)) -> p(p(s(s(s(j(s(p(s(p(s(x1))))))))))) j(s(x1)) -> p(s(s(p(s(f(p(s(p(p(s(x1))))))))))) half(0(x1)) -> 0(s(s(half(p(s(p(s(x1)))))))) half(s(s(x1))) -> s(half(p(p(s(s(x1)))))) rd(0(x1)) -> 0(s(0(0(0(0(s(0(rd(x1))))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (33) MNOCProof (EQUIVALENT) We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R. ---------------------------------------- (34) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(x1)) -> G(p(s(s(x1)))) G(s(x1)) -> J(s(p(s(p(s(x1)))))) J(s(x1)) -> F(p(s(p(p(s(x1)))))) The TRS R consists of the following rules: p(0(x1)) -> 0(s(s(p(x1)))) p(s(x1)) -> x1 p(p(s(x1))) -> p(x1) f(s(x1)) -> p(s(g(p(s(s(x1)))))) g(s(x1)) -> p(p(s(s(s(j(s(p(s(p(s(x1))))))))))) j(s(x1)) -> p(s(s(p(s(f(p(s(p(p(s(x1))))))))))) half(0(x1)) -> 0(s(s(half(p(s(p(s(x1)))))))) half(s(s(x1))) -> s(half(p(p(s(s(x1)))))) rd(0(x1)) -> 0(s(0(0(0(0(s(0(rd(x1))))))))) The set Q consists of the following terms: p(0(x0)) p(s(x0)) f(s(x0)) g(s(x0)) j(s(x0)) half(0(x0)) half(s(s(x0))) rd(0(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (35) 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. ---------------------------------------- (36) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(x1)) -> G(p(s(s(x1)))) G(s(x1)) -> J(s(p(s(p(s(x1)))))) J(s(x1)) -> F(p(s(p(p(s(x1)))))) The TRS R consists of the following rules: p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(p(x1)))) The set Q consists of the following terms: p(0(x0)) p(s(x0)) f(s(x0)) g(s(x0)) j(s(x0)) half(0(x0)) half(s(s(x0))) rd(0(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (37) 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(s(x0)) g(s(x0)) j(s(x0)) half(0(x0)) half(s(s(x0))) rd(0(x0)) ---------------------------------------- (38) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(x1)) -> G(p(s(s(x1)))) G(s(x1)) -> J(s(p(s(p(s(x1)))))) J(s(x1)) -> F(p(s(p(p(s(x1)))))) The TRS R consists of the following rules: p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(p(x1)))) The set Q consists of the following terms: p(0(x0)) p(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (39) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. J(s(x1)) -> F(p(s(p(p(s(x1)))))) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( F_1(x_1) ) = 2x_1 POL( G_1(x_1) ) = 2x_1 POL( J_1(x_1) ) = 2x_1 POL( s_1(x_1) ) = x_1 + 2 POL( p_1(x_1) ) = max{0, x_1 - 2} POL( 0_1(x_1) ) = max{0, -2} The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(p(x1)))) ---------------------------------------- (40) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(x1)) -> G(p(s(s(x1)))) G(s(x1)) -> J(s(p(s(p(s(x1)))))) The TRS R consists of the following rules: p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(p(x1)))) The set Q consists of the following terms: p(0(x0)) p(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (41) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes. ---------------------------------------- (42) TRUE