YES proof of AProVE_07_wiehe01.trs # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) QDPOrderProof [EQUIVALENT, 33 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 0 ms] (14) QDP (15) QDPOrderProof [EQUIVALENT, 12 ms] (16) QDP (17) PisEmptyProof [EQUIVALENT, 0 ms] (18) YES (19) QDP (20) UsableRulesProof [EQUIVALENT, 0 ms] (21) QDP (22) QDPSizeChangeProof [EQUIVALENT, 0 ms] (23) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: times(x, 0) -> 0 times(x, s(y)) -> plus(times(x, y), x) plus(s(x), s(y)) -> s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) plus(s(x), x) -> plus(if(gt(x, x), id(x), id(x)), s(x)) plus(zero, y) -> y plus(id(x), s(y)) -> s(plus(x, if(gt(s(y), y), y, s(y)))) id(x) -> x if(true, x, y) -> x if(false, x, y) -> y not(x) -> if(x, false, true) gt(s(x), zero) -> true gt(zero, y) -> false gt(s(x), s(y)) -> gt(x, y) 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: TIMES(x, s(y)) -> PLUS(times(x, y), x) TIMES(x, s(y)) -> TIMES(x, y) PLUS(s(x), s(y)) -> PLUS(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))) PLUS(s(x), s(y)) -> IF(gt(x, y), x, y) PLUS(s(x), s(y)) -> GT(x, y) PLUS(s(x), s(y)) -> IF(not(gt(x, y)), id(x), id(y)) PLUS(s(x), s(y)) -> NOT(gt(x, y)) PLUS(s(x), s(y)) -> ID(x) PLUS(s(x), s(y)) -> ID(y) PLUS(s(x), x) -> PLUS(if(gt(x, x), id(x), id(x)), s(x)) PLUS(s(x), x) -> IF(gt(x, x), id(x), id(x)) PLUS(s(x), x) -> GT(x, x) PLUS(s(x), x) -> ID(x) PLUS(id(x), s(y)) -> PLUS(x, if(gt(s(y), y), y, s(y))) PLUS(id(x), s(y)) -> IF(gt(s(y), y), y, s(y)) PLUS(id(x), s(y)) -> GT(s(y), y) NOT(x) -> IF(x, false, true) GT(s(x), s(y)) -> GT(x, y) The TRS R consists of the following rules: times(x, 0) -> 0 times(x, s(y)) -> plus(times(x, y), x) plus(s(x), s(y)) -> s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) plus(s(x), x) -> plus(if(gt(x, x), id(x), id(x)), s(x)) plus(zero, y) -> y plus(id(x), s(y)) -> s(plus(x, if(gt(s(y), y), y, s(y)))) id(x) -> x if(true, x, y) -> x if(false, x, y) -> y not(x) -> if(x, false, true) gt(s(x), zero) -> true gt(zero, y) -> false gt(s(x), s(y)) -> gt(x, y) 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 3 SCCs with 13 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: GT(s(x), s(y)) -> GT(x, y) The TRS R consists of the following rules: times(x, 0) -> 0 times(x, s(y)) -> plus(times(x, y), x) plus(s(x), s(y)) -> s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) plus(s(x), x) -> plus(if(gt(x, x), id(x), id(x)), s(x)) plus(zero, y) -> y plus(id(x), s(y)) -> s(plus(x, if(gt(s(y), y), y, s(y)))) id(x) -> x if(true, x, y) -> x if(false, x, y) -> y not(x) -> if(x, false, true) gt(s(x), zero) -> true gt(zero, y) -> false gt(s(x), s(y)) -> gt(x, y) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (6) 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. ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: GT(s(x), s(y)) -> GT(x, y) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) 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: *GT(s(x), s(y)) -> GT(x, y) The graph contains the following edges 1 > 1, 2 > 2 ---------------------------------------- (9) YES ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: PLUS(s(x), x) -> PLUS(if(gt(x, x), id(x), id(x)), s(x)) PLUS(s(x), s(y)) -> PLUS(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))) PLUS(id(x), s(y)) -> PLUS(x, if(gt(s(y), y), y, s(y))) The TRS R consists of the following rules: times(x, 0) -> 0 times(x, s(y)) -> plus(times(x, y), x) plus(s(x), s(y)) -> s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) plus(s(x), x) -> plus(if(gt(x, x), id(x), id(x)), s(x)) plus(zero, y) -> y plus(id(x), s(y)) -> s(plus(x, if(gt(s(y), y), y, s(y)))) id(x) -> x if(true, x, y) -> x if(false, x, y) -> y not(x) -> if(x, false, true) gt(s(x), zero) -> true gt(zero, y) -> false gt(s(x), s(y)) -> gt(x, y) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. PLUS(s(x), s(y)) -> PLUS(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(PLUS(x_1, x_2)) = [[0A]] + [[1A]] * x_1 + [[0A]] * x_2 >>> <<< POL(s(x_1)) = [[0A]] + [[2A]] * x_1 >>> <<< POL(if(x_1, x_2, x_3)) = [[-I]] + [[0A]] * x_1 + [[0A]] * x_2 + [[0A]] * x_3 >>> <<< POL(gt(x_1, x_2)) = [[-I]] + [[0A]] * x_1 + [[-I]] * x_2 >>> <<< POL(id(x_1)) = [[0A]] + [[1A]] * x_1 >>> <<< POL(not(x_1)) = [[0A]] + [[1A]] * x_1 >>> <<< POL(zero) = [[2A]] >>> <<< POL(true) = [[0A]] >>> <<< POL(false) = [[0A]] >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: gt(s(x), zero) -> true gt(zero, y) -> false gt(s(x), s(y)) -> gt(x, y) id(x) -> x if(true, x, y) -> x if(false, x, y) -> y not(x) -> if(x, false, true) ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: PLUS(s(x), x) -> PLUS(if(gt(x, x), id(x), id(x)), s(x)) PLUS(id(x), s(y)) -> PLUS(x, if(gt(s(y), y), y, s(y))) The TRS R consists of the following rules: times(x, 0) -> 0 times(x, s(y)) -> plus(times(x, y), x) plus(s(x), s(y)) -> s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) plus(s(x), x) -> plus(if(gt(x, x), id(x), id(x)), s(x)) plus(zero, y) -> y plus(id(x), s(y)) -> s(plus(x, if(gt(s(y), y), y, s(y)))) id(x) -> x if(true, x, y) -> x if(false, x, y) -> y not(x) -> if(x, false, true) gt(s(x), zero) -> true gt(zero, y) -> false gt(s(x), s(y)) -> gt(x, y) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. PLUS(id(x), s(y)) -> PLUS(x, if(gt(s(y), y), y, s(y))) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( PLUS_2(x_1, x_2) ) = x_1 + 1 POL( if_3(x_1, ..., x_3) ) = 2x_1 + x_2 + x_3 POL( gt_2(x_1, x_2) ) = max{0, -1} POL( s_1(x_1) ) = 2x_1 + 2 POL( zero ) = 0 POL( true ) = 0 POL( false ) = 0 POL( id_1(x_1) ) = x_1 + 1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: gt(s(x), zero) -> true gt(zero, y) -> false gt(s(x), s(y)) -> gt(x, y) id(x) -> x if(true, x, y) -> x if(false, x, y) -> y ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: PLUS(s(x), x) -> PLUS(if(gt(x, x), id(x), id(x)), s(x)) The TRS R consists of the following rules: times(x, 0) -> 0 times(x, s(y)) -> plus(times(x, y), x) plus(s(x), s(y)) -> s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) plus(s(x), x) -> plus(if(gt(x, x), id(x), id(x)), s(x)) plus(zero, y) -> y plus(id(x), s(y)) -> s(plus(x, if(gt(s(y), y), y, s(y)))) id(x) -> x if(true, x, y) -> x if(false, x, y) -> y not(x) -> if(x, false, true) gt(s(x), zero) -> true gt(zero, y) -> false gt(s(x), s(y)) -> gt(x, y) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. PLUS(s(x), x) -> PLUS(if(gt(x, x), id(x), id(x)), s(x)) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( PLUS_2(x_1, x_2) ) = 2x_1 + 2 POL( if_3(x_1, ..., x_3) ) = x_1 + x_2 + x_3 POL( gt_2(x_1, x_2) ) = 0 POL( s_1(x_1) ) = 2x_1 + 1 POL( zero ) = 0 POL( true ) = 0 POL( false ) = 0 POL( id_1(x_1) ) = x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: gt(s(x), zero) -> true gt(zero, y) -> false gt(s(x), s(y)) -> gt(x, y) id(x) -> x if(true, x, y) -> x if(false, x, y) -> y ---------------------------------------- (16) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: times(x, 0) -> 0 times(x, s(y)) -> plus(times(x, y), x) plus(s(x), s(y)) -> s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) plus(s(x), x) -> plus(if(gt(x, x), id(x), id(x)), s(x)) plus(zero, y) -> y plus(id(x), s(y)) -> s(plus(x, if(gt(s(y), y), y, s(y)))) id(x) -> x if(true, x, y) -> x if(false, x, y) -> y not(x) -> if(x, false, true) gt(s(x), zero) -> true gt(zero, y) -> false gt(s(x), s(y)) -> gt(x, y) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (18) YES ---------------------------------------- (19) Obligation: Q DP problem: The TRS P consists of the following rules: TIMES(x, s(y)) -> TIMES(x, y) The TRS R consists of the following rules: times(x, 0) -> 0 times(x, s(y)) -> plus(times(x, y), x) plus(s(x), s(y)) -> s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) plus(s(x), x) -> plus(if(gt(x, x), id(x), id(x)), s(x)) plus(zero, y) -> y plus(id(x), s(y)) -> s(plus(x, if(gt(s(y), y), y, s(y)))) id(x) -> x if(true, x, y) -> x if(false, x, y) -> y not(x) -> if(x, false, true) gt(s(x), zero) -> true gt(zero, y) -> false gt(s(x), s(y)) -> gt(x, y) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (20) 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. ---------------------------------------- (21) Obligation: Q DP problem: The TRS P consists of the following rules: TIMES(x, s(y)) -> TIMES(x, y) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (22) 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: *TIMES(x, s(y)) -> TIMES(x, y) The graph contains the following edges 1 >= 1, 2 > 2 ---------------------------------------- (23) YES