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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 56 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 24 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) QDPSizeChangeProof [EQUIVALENT, 0 ms] (16) YES (17) QDP (18) UsableRulesProof [EQUIVALENT, 0 ms] (19) QDP (20) MRRProof [EQUIVALENT, 12 ms] (21) QDP (22) DependencyGraphProof [EQUIVALENT, 0 ms] (23) AND (24) QDP (25) UsableRulesProof [EQUIVALENT, 0 ms] (26) QDP (27) MRRProof [EQUIVALENT, 5 ms] (28) QDP (29) MNOCProof [EQUIVALENT, 0 ms] (30) QDP (31) QDPOrderProof [EQUIVALENT, 15 ms] (32) QDP (33) PisEmptyProof [EQUIVALENT, 0 ms] (34) YES (35) QDP (36) MRRProof [EQUIVALENT, 22 ms] (37) QDP (38) QDPOrderProof [EQUIVALENT, 0 ms] (39) QDP (40) PisEmptyProof [EQUIVALENT, 0 ms] (41) YES (42) QDP (43) UsableRulesProof [EQUIVALENT, 0 ms] (44) QDP (45) QDPOrderProof [EQUIVALENT, 43 ms] (46) QDP (47) PisEmptyProof [EQUIVALENT, 0 ms] (48) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: log(s(x1)) -> s(log(half(s(x1)))) half(0(x1)) -> 0(s(s(half(x1)))) half(s(0(x1))) -> 0(x1) half(s(s(x1))) -> s(half(p(s(s(x1))))) half(half(s(s(s(s(x1)))))) -> s(s(half(half(x1)))) p(s(s(s(x1)))) -> s(p(s(s(x1)))) s(s(p(s(x1)))) -> 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(half(x_1)) = x_1 POL(log(x_1)) = x_1 POL(p(x_1)) = x_1 POL(s(x_1)) = x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 0(x1) -> x1 ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: log(s(x1)) -> s(log(half(s(x1)))) half(0(x1)) -> 0(s(s(half(x1)))) half(s(0(x1))) -> 0(x1) half(s(s(x1))) -> s(half(p(s(s(x1))))) half(half(s(s(s(s(x1)))))) -> s(s(half(half(x1)))) p(s(s(s(x1)))) -> s(p(s(s(x1)))) s(s(p(s(x1)))) -> 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: LOG(s(x1)) -> S(log(half(s(x1)))) LOG(s(x1)) -> LOG(half(s(x1))) LOG(s(x1)) -> HALF(s(x1)) HALF(0(x1)) -> S(s(half(x1))) HALF(0(x1)) -> S(half(x1)) HALF(0(x1)) -> HALF(x1) HALF(s(s(x1))) -> S(half(p(s(s(x1))))) HALF(s(s(x1))) -> HALF(p(s(s(x1)))) HALF(s(s(x1))) -> P(s(s(x1))) HALF(half(s(s(s(s(x1)))))) -> S(s(half(half(x1)))) HALF(half(s(s(s(s(x1)))))) -> S(half(half(x1))) HALF(half(s(s(s(s(x1)))))) -> HALF(half(x1)) HALF(half(s(s(s(s(x1)))))) -> HALF(x1) P(s(s(s(x1)))) -> S(p(s(s(x1)))) P(s(s(s(x1)))) -> P(s(s(x1))) S(s(p(s(x1)))) -> S(s(x1)) The TRS R consists of the following rules: log(s(x1)) -> s(log(half(s(x1)))) half(0(x1)) -> 0(s(s(half(x1)))) half(s(0(x1))) -> 0(x1) half(s(s(x1))) -> s(half(p(s(s(x1))))) half(half(s(s(s(s(x1)))))) -> s(s(half(half(x1)))) p(s(s(s(x1)))) -> s(p(s(s(x1)))) s(s(p(s(x1)))) -> 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 4 SCCs with 9 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: S(s(p(s(x1)))) -> S(s(x1)) The TRS R consists of the following rules: log(s(x1)) -> s(log(half(s(x1)))) half(0(x1)) -> 0(s(s(half(x1)))) half(s(0(x1))) -> 0(x1) half(s(s(x1))) -> s(half(p(s(s(x1))))) half(half(s(s(s(s(x1)))))) -> s(s(half(half(x1)))) p(s(s(s(x1)))) -> s(p(s(s(x1)))) s(s(p(s(x1)))) -> 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: S(s(p(s(x1)))) -> S(s(x1)) The TRS R consists of the following rules: s(s(p(s(x1)))) -> s(s(x1)) 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: *S(s(p(s(x1)))) -> S(s(x1)) The graph contains the following edges 1 > 1 ---------------------------------------- (11) YES ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: P(s(s(s(x1)))) -> P(s(s(x1))) The TRS R consists of the following rules: log(s(x1)) -> s(log(half(s(x1)))) half(0(x1)) -> 0(s(s(half(x1)))) half(s(0(x1))) -> 0(x1) half(s(s(x1))) -> s(half(p(s(s(x1))))) half(half(s(s(s(s(x1)))))) -> s(s(half(half(x1)))) p(s(s(s(x1)))) -> s(p(s(s(x1)))) s(s(p(s(x1)))) -> s(s(x1)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) 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. ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: P(s(s(s(x1)))) -> P(s(s(x1))) The TRS R consists of the following rules: s(s(p(s(x1)))) -> s(s(x1)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) 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(s(s(s(x1)))) -> P(s(s(x1))) The graph contains the following edges 1 > 1 ---------------------------------------- (16) YES ---------------------------------------- (17) Obligation: Q DP problem: The TRS P consists of the following rules: HALF(s(s(x1))) -> HALF(p(s(s(x1)))) HALF(0(x1)) -> HALF(x1) HALF(half(s(s(s(s(x1)))))) -> HALF(half(x1)) HALF(half(s(s(s(s(x1)))))) -> HALF(x1) The TRS R consists of the following rules: log(s(x1)) -> s(log(half(s(x1)))) half(0(x1)) -> 0(s(s(half(x1)))) half(s(0(x1))) -> 0(x1) half(s(s(x1))) -> s(half(p(s(s(x1))))) half(half(s(s(s(s(x1)))))) -> s(s(half(half(x1)))) p(s(s(s(x1)))) -> s(p(s(s(x1)))) s(s(p(s(x1)))) -> s(s(x1)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (18) 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. ---------------------------------------- (19) Obligation: Q DP problem: The TRS P consists of the following rules: HALF(s(s(x1))) -> HALF(p(s(s(x1)))) HALF(0(x1)) -> HALF(x1) HALF(half(s(s(s(s(x1)))))) -> HALF(half(x1)) HALF(half(s(s(s(s(x1)))))) -> HALF(x1) The TRS R consists of the following rules: half(0(x1)) -> 0(s(s(half(x1)))) half(s(0(x1))) -> 0(x1) half(s(s(x1))) -> s(half(p(s(s(x1))))) half(half(s(s(s(s(x1)))))) -> s(s(half(half(x1)))) s(s(p(s(x1)))) -> s(s(x1)) p(s(s(s(x1)))) -> s(p(s(s(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (20) 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: HALF(0(x1)) -> HALF(x1) Strictly oriented rules of the TRS R: half(0(x1)) -> 0(s(s(half(x1)))) half(s(0(x1))) -> 0(x1) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = 1 + x_1 POL(HALF(x_1)) = 2*x_1 POL(half(x_1)) = 2*x_1 POL(p(x_1)) = x_1 POL(s(x_1)) = x_1 ---------------------------------------- (21) Obligation: Q DP problem: The TRS P consists of the following rules: HALF(s(s(x1))) -> HALF(p(s(s(x1)))) HALF(half(s(s(s(s(x1)))))) -> HALF(half(x1)) HALF(half(s(s(s(s(x1)))))) -> HALF(x1) The TRS R consists of the following rules: half(s(s(x1))) -> s(half(p(s(s(x1))))) half(half(s(s(s(s(x1)))))) -> s(s(half(half(x1)))) s(s(p(s(x1)))) -> s(s(x1)) p(s(s(s(x1)))) -> s(p(s(s(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (22) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs. ---------------------------------------- (23) Complex Obligation (AND) ---------------------------------------- (24) Obligation: Q DP problem: The TRS P consists of the following rules: HALF(s(s(x1))) -> HALF(p(s(s(x1)))) The TRS R consists of the following rules: half(s(s(x1))) -> s(half(p(s(s(x1))))) half(half(s(s(s(s(x1)))))) -> s(s(half(half(x1)))) s(s(p(s(x1)))) -> s(s(x1)) p(s(s(s(x1)))) -> s(p(s(s(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (25) 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. ---------------------------------------- (26) Obligation: Q DP problem: The TRS P consists of the following rules: HALF(s(s(x1))) -> HALF(p(s(s(x1)))) The TRS R consists of the following rules: s(s(p(s(x1)))) -> s(s(x1)) p(s(s(s(x1)))) -> s(p(s(s(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (27) 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: s(s(p(s(x1)))) -> s(s(x1)) Used ordering: Polynomial interpretation [POLO]: POL(HALF(x_1)) = x_1 POL(p(x_1)) = x_1 POL(s(x_1)) = 2 + 2*x_1 ---------------------------------------- (28) Obligation: Q DP problem: The TRS P consists of the following rules: HALF(s(s(x1))) -> HALF(p(s(s(x1)))) The TRS R consists of the following rules: p(s(s(s(x1)))) -> s(p(s(s(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (29) MNOCProof (EQUIVALENT) We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R. ---------------------------------------- (30) Obligation: Q DP problem: The TRS P consists of the following rules: HALF(s(s(x1))) -> HALF(p(s(s(x1)))) The TRS R consists of the following rules: p(s(s(s(x1)))) -> s(p(s(s(x1)))) The set Q consists of the following terms: p(s(s(s(x0)))) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (31) 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(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) ) = max{0, 2x_1 - 2} POL( p_1(x_1) ) = max{0, x_1 - 2} POL( s_1(x_1) ) = 2x_1 + 1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: p(s(s(s(x1)))) -> s(p(s(s(x1)))) ---------------------------------------- (32) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: p(s(s(s(x1)))) -> s(p(s(s(x1)))) The set Q consists of the following terms: p(s(s(s(x0)))) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (33) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (34) YES ---------------------------------------- (35) Obligation: Q DP problem: The TRS P consists of the following rules: HALF(half(s(s(s(s(x1)))))) -> HALF(x1) HALF(half(s(s(s(s(x1)))))) -> HALF(half(x1)) The TRS R consists of the following rules: half(s(s(x1))) -> s(half(p(s(s(x1))))) half(half(s(s(s(s(x1)))))) -> s(s(half(half(x1)))) s(s(p(s(x1)))) -> s(s(x1)) p(s(s(s(x1)))) -> s(p(s(s(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (36) 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: HALF(half(s(s(s(s(x1)))))) -> HALF(x1) Used ordering: Polynomial interpretation [POLO]: POL(HALF(x_1)) = x_1 POL(half(x_1)) = 1 + x_1 POL(p(x_1)) = x_1 POL(s(x_1)) = x_1 ---------------------------------------- (37) Obligation: Q DP problem: The TRS P consists of the following rules: HALF(half(s(s(s(s(x1)))))) -> HALF(half(x1)) The TRS R consists of the following rules: half(s(s(x1))) -> s(half(p(s(s(x1))))) half(half(s(s(s(s(x1)))))) -> s(s(half(half(x1)))) s(s(p(s(x1)))) -> s(s(x1)) p(s(s(s(x1)))) -> s(p(s(s(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (38) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. HALF(half(s(s(s(s(x1)))))) -> HALF(half(x1)) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( HALF_1(x_1) ) = max{0, 2x_1 - 2} POL( half_1(x_1) ) = max{0, 2x_1 - 2} POL( s_1(x_1) ) = x_1 + 2 POL( p_1(x_1) ) = max{0, x_1 - 2} The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: half(s(s(x1))) -> s(half(p(s(s(x1))))) half(half(s(s(s(s(x1)))))) -> s(s(half(half(x1)))) s(s(p(s(x1)))) -> s(s(x1)) p(s(s(s(x1)))) -> s(p(s(s(x1)))) ---------------------------------------- (39) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: half(s(s(x1))) -> s(half(p(s(s(x1))))) half(half(s(s(s(s(x1)))))) -> s(s(half(half(x1)))) s(s(p(s(x1)))) -> s(s(x1)) p(s(s(s(x1)))) -> s(p(s(s(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (40) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (41) YES ---------------------------------------- (42) Obligation: Q DP problem: The TRS P consists of the following rules: LOG(s(x1)) -> LOG(half(s(x1))) The TRS R consists of the following rules: log(s(x1)) -> s(log(half(s(x1)))) half(0(x1)) -> 0(s(s(half(x1)))) half(s(0(x1))) -> 0(x1) half(s(s(x1))) -> s(half(p(s(s(x1))))) half(half(s(s(s(s(x1)))))) -> s(s(half(half(x1)))) p(s(s(s(x1)))) -> s(p(s(s(x1)))) s(s(p(s(x1)))) -> s(s(x1)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (43) 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. ---------------------------------------- (44) Obligation: Q DP problem: The TRS P consists of the following rules: LOG(s(x1)) -> LOG(half(s(x1))) The TRS R consists of the following rules: s(s(p(s(x1)))) -> s(s(x1)) half(0(x1)) -> 0(s(s(half(x1)))) half(s(0(x1))) -> 0(x1) half(s(s(x1))) -> s(half(p(s(s(x1))))) half(half(s(s(s(s(x1)))))) -> s(s(half(half(x1)))) p(s(s(s(x1)))) -> s(p(s(s(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (45) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. LOG(s(x1)) -> LOG(half(s(x1))) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( LOG_1(x_1) ) = x_1 + 2 POL( s_1(x_1) ) = x_1 + 1 POL( p_1(x_1) ) = max{0, x_1 - 1} POL( half_1(x_1) ) = max{0, x_1 - 1} 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: s(s(p(s(x1)))) -> s(s(x1)) half(0(x1)) -> 0(s(s(half(x1)))) half(s(0(x1))) -> 0(x1) half(s(s(x1))) -> s(half(p(s(s(x1))))) half(half(s(s(s(s(x1)))))) -> s(s(half(half(x1)))) p(s(s(s(x1)))) -> s(p(s(s(x1)))) ---------------------------------------- (46) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: s(s(p(s(x1)))) -> s(s(x1)) half(0(x1)) -> 0(s(s(half(x1)))) half(s(0(x1))) -> 0(x1) half(s(s(x1))) -> s(half(p(s(s(x1))))) half(half(s(s(s(s(x1)))))) -> s(s(half(half(x1)))) p(s(s(s(x1)))) -> s(p(s(s(x1)))) Q is empty. 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