/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 4 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) ATransformationProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) ATransformationProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPSizeChangeProof [EQUIVALENT, 0 ms] (18) YES (19) QDP (20) QDPApplicativeOrderProof [EQUIVALENT, 28 ms] (21) QDP (22) PisEmptyProof [EQUIVALENT, 0 ms] (23) YES (24) QDP (25) QDPSizeChangeProof [EQUIVALENT, 0 ms] (26) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(minus, x), 0) -> x app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y) app(double, 0) -> 0 app(double, app(s, x)) -> app(s, app(s, app(double, x))) app(app(plus, 0), y) -> y app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y)) app(app(plus, app(s, x)), y) -> app(app(plus, x), app(s, y)) app(app(plus, app(s, x)), y) -> app(s, app(app(plus, app(app(minus, x), y)), app(double, y))) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) 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: APP(app(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y) APP(app(minus, app(s, x)), app(s, y)) -> APP(minus, x) APP(double, app(s, x)) -> APP(s, app(s, app(double, x))) APP(double, app(s, x)) -> APP(s, app(double, x)) APP(double, app(s, x)) -> APP(double, x) APP(app(plus, app(s, x)), y) -> APP(s, app(app(plus, x), y)) APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y) APP(app(plus, app(s, x)), y) -> APP(plus, x) APP(app(plus, app(s, x)), y) -> APP(app(plus, x), app(s, y)) APP(app(plus, app(s, x)), y) -> APP(s, y) APP(app(plus, app(s, x)), y) -> APP(s, app(app(plus, app(app(minus, x), y)), app(double, y))) APP(app(plus, app(s, x)), y) -> APP(app(plus, app(app(minus, x), y)), app(double, y)) APP(app(plus, app(s, x)), y) -> APP(plus, app(app(minus, x), y)) APP(app(plus, app(s, x)), y) -> APP(app(minus, x), y) APP(app(plus, app(s, x)), y) -> APP(minus, x) APP(app(plus, app(s, x)), y) -> APP(double, y) APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs)) APP(app(map, f), app(app(cons, x), xs)) -> APP(cons, app(f, x)) APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(f, x)), f), x), xs) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(filter2, app(f, x)), f), x) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(filter2, app(f, x)), f) APP(app(filter, f), app(app(cons, x), xs)) -> APP(filter2, app(f, x)) APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(app(app(filter2, true), f), x), xs) -> APP(app(cons, x), app(app(filter, f), xs)) APP(app(app(app(filter2, true), f), x), xs) -> APP(cons, x) APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) APP(app(app(app(filter2, true), f), x), xs) -> APP(filter, f) APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) APP(app(app(app(filter2, false), f), x), xs) -> APP(filter, f) The TRS R consists of the following rules: app(app(minus, x), 0) -> x app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y) app(double, 0) -> 0 app(double, app(s, x)) -> app(s, app(s, app(double, x))) app(app(plus, 0), y) -> y app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y)) app(app(plus, app(s, x)), y) -> app(app(plus, x), app(s, y)) app(app(plus, app(s, x)), y) -> app(s, app(app(plus, app(app(minus, x), y)), app(double, y))) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) 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 20 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: APP(double, app(s, x)) -> APP(double, x) The TRS R consists of the following rules: app(app(minus, x), 0) -> x app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y) app(double, 0) -> 0 app(double, app(s, x)) -> app(s, app(s, app(double, x))) app(app(plus, 0), y) -> y app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y)) app(app(plus, app(s, x)), y) -> app(app(plus, x), app(s, y)) app(app(plus, app(s, x)), y) -> app(s, app(app(plus, app(app(minus, x), y)), app(double, y))) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) 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: APP(double, app(s, x)) -> APP(double, x) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) ATransformationProof (EQUIVALENT) We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: double(s(x)) -> double(x) 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: *double(s(x)) -> double(x) The graph contains the following edges 1 > 1 ---------------------------------------- (11) YES ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y) The TRS R consists of the following rules: app(app(minus, x), 0) -> x app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y) app(double, 0) -> 0 app(double, app(s, x)) -> app(s, app(s, app(double, x))) app(app(plus, 0), y) -> y app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y)) app(app(plus, app(s, x)), y) -> app(app(plus, x), app(s, y)) app(app(plus, app(s, x)), y) -> app(s, app(app(plus, app(app(minus, x), y)), app(double, y))) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) 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: APP(app(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) ATransformationProof (EQUIVALENT) We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: minus(s(x), s(y)) -> minus(x, y) 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: *minus(s(x), s(y)) -> minus(x, y) The graph contains the following edges 1 > 1, 2 > 2 ---------------------------------------- (18) YES ---------------------------------------- (19) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(plus, app(s, x)), y) -> APP(app(plus, x), app(s, y)) APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y) APP(app(plus, app(s, x)), y) -> APP(app(plus, app(app(minus, x), y)), app(double, y)) The TRS R consists of the following rules: app(app(minus, x), 0) -> x app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y) app(double, 0) -> 0 app(double, app(s, x)) -> app(s, app(s, app(double, x))) app(app(plus, 0), y) -> y app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y)) app(app(plus, app(s, x)), y) -> app(app(plus, x), app(s, y)) app(app(plus, app(s, x)), y) -> app(s, app(app(plus, app(app(minus, x), y)), app(double, y))) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (20) QDPApplicativeOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04].Here, we combined the reduction pair processor [LPAR04,JAR06] with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem. The a-transformed P is (plus1(s(x), y),plus1(x, s(y))) (plus1(s(x), y),plus1(x, y)) (plus1(s(x), y),plus1(minus(x, y), double(y))) The a-transformed usable rules are double(0) -> 0 minus(x, 0) -> x minus(s(x), s(y)) -> minus(x, y) The following pairs can be oriented strictly and are deleted. APP(app(plus, app(s, x)), y) -> APP(app(plus, x), app(s, y)) APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y) APP(app(plus, app(s, x)), y) -> APP(app(plus, app(app(minus, x), y)), app(double, y)) The remaining pairs can at least be oriented weakly. none Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( double_1(x_1) ) = x_1 + 2 POL( 0 ) = 2 POL( s_1(x_1) ) = x_1 + 2 POL( minus_2(x_1, x_2) ) = x_1 + 1 POL( plus1_2(x_1, x_2) ) = x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: app(double, 0) -> 0 app(app(minus, x), 0) -> x app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y) ---------------------------------------- (21) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: app(app(minus, x), 0) -> x app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y) app(double, 0) -> 0 app(double, app(s, x)) -> app(s, app(s, app(double, x))) app(app(plus, 0), y) -> y app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y)) app(app(plus, app(s, x)), y) -> app(app(plus, x), app(s, y)) app(app(plus, app(s, x)), y) -> app(s, app(app(plus, app(app(minus, x), y)), app(double, y))) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (22) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (23) YES ---------------------------------------- (24) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(f, x)), f), x), xs) APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) The TRS R consists of the following rules: app(app(minus, x), 0) -> x app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y) app(double, 0) -> 0 app(double, app(s, x)) -> app(s, app(s, app(double, x))) app(app(plus, 0), y) -> y app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y)) app(app(plus, app(s, x)), y) -> app(app(plus, x), app(s, y)) app(app(plus, app(s, x)), y) -> app(s, app(app(plus, app(app(minus, x), y)), app(double, y))) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (25) 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: *APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) The graph contains the following edges 1 > 1, 2 > 2 *APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) The graph contains the following edges 1 > 1, 2 > 2 *APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) The graph contains the following edges 1 >= 1, 2 > 2 *APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(f, x)), f), x), xs) The graph contains the following edges 2 > 2 *APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) The graph contains the following edges 2 >= 2 *APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) The graph contains the following edges 2 >= 2 ---------------------------------------- (26) YES