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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 70 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 47 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 1 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) MNOCProof [EQUIVALENT, 0 ms] (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QReductionProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPOrderProof [EQUIVALENT, 18 ms] (20) QDP (21) PisEmptyProof [EQUIVALENT, 0 ms] (22) YES (23) QDP (24) MNOCProof [EQUIVALENT, 0 ms] (25) QDP (26) UsableRulesProof [EQUIVALENT, 0 ms] (27) QDP (28) QReductionProof [EQUIVALENT, 0 ms] (29) QDP (30) QDPOrderProof [EQUIVALENT, 104 ms] (31) QDP (32) PisEmptyProof [EQUIVALENT, 0 ms] (33) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: sq(0(x1)) -> p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1))))))))))))))))) sq(s(x1)) -> s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))))))) twice(0(x1)) -> p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))))))))) twice(s(x1)) -> p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1))))))))))))))) p(p(s(x1))) -> p(x1) p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(s(s(s(s(s(s(s(s(s(x1)))))))))))) 0(x1) -> x1 Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = 1 + x_1 POL(p(x_1)) = x_1 POL(s(x_1)) = x_1 POL(sq(x_1)) = 1 + x_1 POL(twice(x_1)) = x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: sq(0(x1)) -> p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1))))))))))))))))) 0(x1) -> x1 ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: sq(s(x1)) -> s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))))))) twice(0(x1)) -> p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))))))))) twice(s(x1)) -> p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1))))))))))))))) p(p(s(x1))) -> p(x1) p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(s(s(s(s(s(s(s(s(s(x1)))))))))))) Q is empty. ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: SQ(s(x1)) -> P(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))) SQ(s(x1)) -> P(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))) SQ(s(x1)) -> P(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))) SQ(s(x1)) -> P(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))) SQ(s(x1)) -> TWICE(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))) SQ(s(x1)) -> P(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))) SQ(s(x1)) -> P(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))) SQ(s(x1)) -> P(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))) SQ(s(x1)) -> P(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))) SQ(s(x1)) -> P(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))) SQ(s(x1)) -> SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))) SQ(s(x1)) -> P(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))) SQ(s(x1)) -> P(p(p(p(p(s(s(s(s(s(s(x1))))))))))) SQ(s(x1)) -> P(p(p(p(s(s(s(s(s(s(x1)))))))))) SQ(s(x1)) -> P(p(p(s(s(s(s(s(s(x1))))))))) SQ(s(x1)) -> P(p(s(s(s(s(s(s(x1)))))))) SQ(s(x1)) -> P(s(s(s(s(s(s(x1))))))) TWICE(0(x1)) -> P(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))))))))) TWICE(0(x1)) -> P(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))) TWICE(0(x1)) -> P(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))))))) TWICE(0(x1)) -> P(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))) TWICE(0(x1)) -> P(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))) TWICE(0(x1)) -> P(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))) TWICE(0(x1)) -> P(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))) TWICE(0(x1)) -> P(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))) TWICE(0(x1)) -> P(p(s(s(p(s(p(s(p(s(x1)))))))))) TWICE(0(x1)) -> P(s(s(p(s(p(s(p(s(x1))))))))) TWICE(0(x1)) -> P(s(p(s(p(s(x1)))))) TWICE(0(x1)) -> P(s(p(s(x1)))) TWICE(0(x1)) -> P(s(x1)) TWICE(s(x1)) -> P(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1))))))))))))))) TWICE(s(x1)) -> P(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))) TWICE(s(x1)) -> P(p(s(s(s(twice(p(s(p(s(x1)))))))))) TWICE(s(x1)) -> P(s(s(s(twice(p(s(p(s(x1))))))))) TWICE(s(x1)) -> TWICE(p(s(p(s(x1))))) TWICE(s(x1)) -> P(s(p(s(x1)))) TWICE(s(x1)) -> P(s(x1)) P(p(s(x1))) -> P(x1) The TRS R consists of the following rules: sq(s(x1)) -> s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))))))) twice(0(x1)) -> p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))))))))) twice(s(x1)) -> p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1))))))))))))))) p(p(s(x1))) -> p(x1) p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(s(s(s(s(s(s(s(s(s(x1)))))))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 35 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: P(p(s(x1))) -> P(x1) The TRS R consists of the following rules: sq(s(x1)) -> s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))))))) twice(0(x1)) -> p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))))))))) twice(s(x1)) -> p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1))))))))))))))) p(p(s(x1))) -> p(x1) p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(s(s(s(s(s(s(s(s(s(x1)))))))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) 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. ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: P(p(s(x1))) -> P(x1) R is empty. Q is empty. 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: *P(p(s(x1))) -> P(x1) The graph contains the following edges 1 > 1 ---------------------------------------- (11) YES ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: TWICE(s(x1)) -> TWICE(p(s(p(s(x1))))) The TRS R consists of the following rules: sq(s(x1)) -> s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))))))) twice(0(x1)) -> p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))))))))) twice(s(x1)) -> p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1))))))))))))))) p(p(s(x1))) -> p(x1) p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(s(s(s(s(s(s(s(s(s(x1)))))))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) MNOCProof (EQUIVALENT) We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R. ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: TWICE(s(x1)) -> TWICE(p(s(p(s(x1))))) The TRS R consists of the following rules: sq(s(x1)) -> s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))))))) twice(0(x1)) -> p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))))))))) twice(s(x1)) -> p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1))))))))))))))) p(p(s(x1))) -> p(x1) p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(s(s(s(s(s(s(s(s(s(x1)))))))))))) The set Q consists of the following terms: sq(s(x0)) twice(0(x0)) twice(s(x0)) p(s(x0)) p(0(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) 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. ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: TWICE(s(x1)) -> TWICE(p(s(p(s(x1))))) The TRS R consists of the following rules: p(s(x1)) -> x1 The set Q consists of the following terms: sq(s(x0)) twice(0(x0)) twice(s(x0)) p(s(x0)) p(0(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) 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]. sq(s(x0)) twice(0(x0)) twice(s(x0)) ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: TWICE(s(x1)) -> TWICE(p(s(p(s(x1))))) The TRS R consists of the following rules: p(s(x1)) -> x1 The set Q consists of the following terms: p(s(x0)) p(0(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. TWICE(s(x1)) -> TWICE(p(s(p(s(x1))))) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( TWICE_1(x_1) ) = x_1 + 2 POL( p_1(x_1) ) = max{0, x_1 - 1} POL( s_1(x_1) ) = x_1 + 1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: p(s(x1)) -> x1 ---------------------------------------- (20) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: p(s(x1)) -> x1 The set Q consists of the following terms: p(s(x0)) p(0(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (22) YES ---------------------------------------- (23) Obligation: Q DP problem: The TRS P consists of the following rules: SQ(s(x1)) -> SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))) The TRS R consists of the following rules: sq(s(x1)) -> s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))))))) twice(0(x1)) -> p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))))))))) twice(s(x1)) -> p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1))))))))))))))) p(p(s(x1))) -> p(x1) p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(s(s(s(s(s(s(s(s(s(x1)))))))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (24) MNOCProof (EQUIVALENT) We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R. ---------------------------------------- (25) Obligation: Q DP problem: The TRS P consists of the following rules: SQ(s(x1)) -> SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))) The TRS R consists of the following rules: sq(s(x1)) -> s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))))))) twice(0(x1)) -> p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))))))))) twice(s(x1)) -> p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1))))))))))))))) p(p(s(x1))) -> p(x1) p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(s(s(s(s(s(s(s(s(s(x1)))))))))))) The set Q consists of the following terms: sq(s(x0)) twice(0(x0)) twice(s(x0)) p(s(x0)) p(0(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (26) 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. ---------------------------------------- (27) Obligation: Q DP problem: The TRS P consists of the following rules: SQ(s(x1)) -> SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))) The TRS R consists of the following rules: p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(s(s(s(s(s(s(s(s(s(x1)))))))))))) The set Q consists of the following terms: sq(s(x0)) twice(0(x0)) twice(s(x0)) p(s(x0)) p(0(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (28) 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]. sq(s(x0)) twice(0(x0)) twice(s(x0)) ---------------------------------------- (29) Obligation: Q DP problem: The TRS P consists of the following rules: SQ(s(x1)) -> SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))) The TRS R consists of the following rules: p(s(x1)) -> x1 p(0(x1)) -> 0(s(s(s(s(s(s(s(s(s(s(s(x1)))))))))))) The set Q consists of the following terms: p(s(x0)) p(0(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (30) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. SQ(s(x1)) -> SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( SQ_1(x_1) ) = 2x_1 + 2 POL( p_1(x_1) ) = max{0, x_1 - 2} POL( s_1(x_1) ) = 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(s(s(s(s(s(s(s(s(s(x1)))))))))))) ---------------------------------------- (31) 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(s(s(s(s(s(s(s(s(s(x1)))))))))))) The set Q consists of the following terms: p(s(x0)) p(0(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (32) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (33) YES