/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.pl /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.pl # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern p(g) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToDTProblemTransformerProof [SOUND, 0 ms] (2) TRIPLES (3) TriplesToPiDPProof [SOUND, 0 ms] (4) PiDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) PiDP (7) PiDPToQDPProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Clauses: p(X) :- ','(q(f(Y)), p(Y)). p(g(X)) :- p(X). q(g(Y)). Query: p(g) ---------------------------------------- (1) PrologToDTProblemTransformerProof (SOUND) Built DT problem from termination graph DT10. { "root": 14, "program": { "directives": [], "clauses": [ [ "(p X)", "(',' (q (f Y)) (p Y))" ], [ "(p (g X))", "(p X)" ], [ "(q (g Y))", null ] ] }, "graph": { "nodes": { "110": { "goal": [ { "clause": 0, "scope": 1, "term": "(p T1)" }, { "clause": 1, "scope": 1, "term": "(p T1)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": ["T1"], "free": [], "exprvars": [] } }, "122": { "goal": [{ "clause": 1, "scope": 1, "term": "(p T3)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": ["T3"], "free": [], "exprvars": [] } }, "14": { "goal": [{ "clause": -1, "scope": -1, "term": "(p T1)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": ["T1"], "free": [], "exprvars": [] } }, "115": { "goal": [ { "clause": -1, "scope": -1, "term": "(',' (q (f X3)) (p X3))" }, { "clause": 1, "scope": 1, "term": "(p T3)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": ["T3"], "free": ["X3"], "exprvars": [] } }, "118": { "goal": [ { "clause": 2, "scope": 2, "term": "(',' (q (f X3)) (p X3))" }, { "clause": -1, "scope": 2, "term": null }, { "clause": 1, "scope": 1, "term": "(p T3)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": ["T3"], "free": ["X3"], "exprvars": [] } }, "129": { "goal": [{ "clause": -1, "scope": -1, "term": "(p T6)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": ["T6"], "free": [], "exprvars": [] } }, "type": "Nodes", "130": { "goal": [], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "120": { "goal": [ { "clause": -1, "scope": 2, "term": null }, { "clause": 1, "scope": 1, "term": "(p T3)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": ["T3"], "free": [], "exprvars": [] } } }, "edges": [ { "from": 14, "to": 110, "label": "CASE" }, { "from": 110, "to": 115, "label": "ONLY EVAL with clause\np(X2) :- ','(q(f(X3)), p(X3)).\nand substitutionT1 -> T3,\nX2 -> T3" }, { "from": 115, "to": 118, "label": "CASE" }, { "from": 118, "to": 120, "label": "BACKTRACK\nfor clause: q(g(Y))because of non-unification" }, { "from": 120, "to": 122, "label": "FAILURE" }, { "from": 122, "to": 129, "label": "EVAL with clause\np(g(X7)) :- p(X7).\nand substitutionX7 -> T6,\nT3 -> g(T6)" }, { "from": 122, "to": 130, "label": "EVAL-BACKTRACK" }, { "from": 129, "to": 14, "label": "INSTANCE with matching:\nT1 -> T6" } ], "type": "Graph" } } ---------------------------------------- (2) Obligation: Triples: pA(g(X1)) :- pA(X1). Clauses: pcA(g(X1)) :- pcA(X1). Afs: pA(x1) = pA(x1) ---------------------------------------- (3) TriplesToPiDPProof (SOUND) We use the technique of [DT09]. With regard to the inferred argument filtering the predicates were used in the following modes: pA_in_1: (b) Transforming TRIPLES into the following Term Rewriting System: Pi DP problem: The TRS P consists of the following rules: PA_IN_G(g(X1)) -> U1_G(X1, pA_in_g(X1)) PA_IN_G(g(X1)) -> PA_IN_G(X1) R is empty. Pi is empty. We have to consider all (P,R,Pi)-chains Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES ---------------------------------------- (4) Obligation: Pi DP problem: The TRS P consists of the following rules: PA_IN_G(g(X1)) -> U1_G(X1, pA_in_g(X1)) PA_IN_G(g(X1)) -> PA_IN_G(X1) R is empty. Pi is empty. We have to consider all (P,R,Pi)-chains ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node. ---------------------------------------- (6) Obligation: Pi DP problem: The TRS P consists of the following rules: PA_IN_G(g(X1)) -> PA_IN_G(X1) R is empty. Pi is empty. We have to consider all (P,R,Pi)-chains ---------------------------------------- (7) PiDPToQDPProof (EQUIVALENT) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: PA_IN_G(g(X1)) -> PA_IN_G(X1) R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (9) 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: *PA_IN_G(g(X1)) -> PA_IN_G(X1) The graph contains the following edges 1 > 1 ---------------------------------------- (10) YES