/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.pl /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- MAYBE proof of /export/starexec/sandbox/benchmark/theBenchmark.pl # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern p(a) w.r.t. the given Prolog program could not be shown: (0) Prolog (1) CutEliminatorProof [SOUND, 0 ms] (2) Prolog (3) PrologToPiTRSProof [SOUND, 0 ms] (4) PiTRS (5) DependencyPairsProof [EQUIVALENT, 0 ms] (6) PiDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) PiDP (9) UsableRulesProof [EQUIVALENT, 0 ms] (10) PiDP (11) PiDPToQDPProof [SOUND, 0 ms] (12) QDP (13) NonTerminationLoopProof [COMPLETE, 0 ms] (14) NO (15) PrologToPiTRSProof [SOUND, 0 ms] (16) PiTRS (17) DependencyPairsProof [EQUIVALENT, 0 ms] (18) PiDP (19) DependencyGraphProof [EQUIVALENT, 0 ms] (20) PiDP (21) UsableRulesProof [EQUIVALENT, 0 ms] (22) PiDP (23) PiDPToQDPProof [SOUND, 0 ms] (24) QDP (25) NonTerminationLoopProof [COMPLETE, 0 ms] (26) NO (27) PrologToTRSTransformerProof [SOUND, 0 ms] (28) QTRS (29) QTRSRRRProof [EQUIVALENT, 32 ms] (30) QTRS (31) Overlay + Local Confluence [EQUIVALENT, 0 ms] (32) QTRS (33) DependencyPairsProof [EQUIVALENT, 0 ms] (34) QDP (35) UsableRulesProof [EQUIVALENT, 0 ms] (36) QDP (37) QReductionProof [EQUIVALENT, 0 ms] (38) QDP (39) NonTerminationLoopProof [COMPLETE, 0 ms] (40) NO (41) PrologToDTProblemTransformerProof [SOUND, 0 ms] (42) TRIPLES (43) TriplesToPiDPProof [SOUND, 0 ms] (44) PiDP (45) DependencyGraphProof [EQUIVALENT, 0 ms] (46) PiDP (47) PiDPToQDPProof [SOUND, 8 ms] (48) QDP (49) NonTerminationLoopProof [COMPLETE, 0 ms] (50) NO ---------------------------------------- (0) Obligation: Clauses: p(X) :- ','(q(a), !). p(X) :- p(X). q(b). Query: p(a) ---------------------------------------- (1) CutEliminatorProof (SOUND) Eliminated all cuts by simply ignoring them[PROLOG]. ---------------------------------------- (2) Obligation: Clauses: p(X) :- q(a). p(X) :- p(X). q(b). Query: p(a) ---------------------------------------- (3) PrologToPiTRSProof (SOUND) We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: p_in_1: (f) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: p_in_a(X) -> U1_a(X, q_in_g(a)) q_in_g(b) -> q_out_g(b) U1_a(X, q_out_g(a)) -> p_out_a(X) p_in_a(X) -> U2_a(X, p_in_a(X)) U2_a(X, p_out_a(X)) -> p_out_a(X) The argument filtering Pi contains the following mapping: p_in_a(x1) = p_in_a U1_a(x1, x2) = U1_a(x2) q_in_g(x1) = q_in_g(x1) b = b q_out_g(x1) = q_out_g a = a p_out_a(x1) = p_out_a U2_a(x1, x2) = U2_a(x2) Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (4) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: p_in_a(X) -> U1_a(X, q_in_g(a)) q_in_g(b) -> q_out_g(b) U1_a(X, q_out_g(a)) -> p_out_a(X) p_in_a(X) -> U2_a(X, p_in_a(X)) U2_a(X, p_out_a(X)) -> p_out_a(X) The argument filtering Pi contains the following mapping: p_in_a(x1) = p_in_a U1_a(x1, x2) = U1_a(x2) q_in_g(x1) = q_in_g(x1) b = b q_out_g(x1) = q_out_g a = a p_out_a(x1) = p_out_a U2_a(x1, x2) = U2_a(x2) ---------------------------------------- (5) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem: Pi DP problem: The TRS P consists of the following rules: P_IN_A(X) -> U1_A(X, q_in_g(a)) P_IN_A(X) -> Q_IN_G(a) P_IN_A(X) -> U2_A(X, p_in_a(X)) P_IN_A(X) -> P_IN_A(X) The TRS R consists of the following rules: p_in_a(X) -> U1_a(X, q_in_g(a)) q_in_g(b) -> q_out_g(b) U1_a(X, q_out_g(a)) -> p_out_a(X) p_in_a(X) -> U2_a(X, p_in_a(X)) U2_a(X, p_out_a(X)) -> p_out_a(X) The argument filtering Pi contains the following mapping: p_in_a(x1) = p_in_a U1_a(x1, x2) = U1_a(x2) q_in_g(x1) = q_in_g(x1) b = b q_out_g(x1) = q_out_g a = a p_out_a(x1) = p_out_a U2_a(x1, x2) = U2_a(x2) P_IN_A(x1) = P_IN_A U1_A(x1, x2) = U1_A(x2) Q_IN_G(x1) = Q_IN_G(x1) U2_A(x1, x2) = U2_A(x2) We have to consider all (P,R,Pi)-chains ---------------------------------------- (6) Obligation: Pi DP problem: The TRS P consists of the following rules: P_IN_A(X) -> U1_A(X, q_in_g(a)) P_IN_A(X) -> Q_IN_G(a) P_IN_A(X) -> U2_A(X, p_in_a(X)) P_IN_A(X) -> P_IN_A(X) The TRS R consists of the following rules: p_in_a(X) -> U1_a(X, q_in_g(a)) q_in_g(b) -> q_out_g(b) U1_a(X, q_out_g(a)) -> p_out_a(X) p_in_a(X) -> U2_a(X, p_in_a(X)) U2_a(X, p_out_a(X)) -> p_out_a(X) The argument filtering Pi contains the following mapping: p_in_a(x1) = p_in_a U1_a(x1, x2) = U1_a(x2) q_in_g(x1) = q_in_g(x1) b = b q_out_g(x1) = q_out_g a = a p_out_a(x1) = p_out_a U2_a(x1, x2) = U2_a(x2) P_IN_A(x1) = P_IN_A U1_A(x1, x2) = U1_A(x2) Q_IN_G(x1) = Q_IN_G(x1) U2_A(x1, x2) = U2_A(x2) We have to consider all (P,R,Pi)-chains ---------------------------------------- (7) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes. ---------------------------------------- (8) Obligation: Pi DP problem: The TRS P consists of the following rules: P_IN_A(X) -> P_IN_A(X) The TRS R consists of the following rules: p_in_a(X) -> U1_a(X, q_in_g(a)) q_in_g(b) -> q_out_g(b) U1_a(X, q_out_g(a)) -> p_out_a(X) p_in_a(X) -> U2_a(X, p_in_a(X)) U2_a(X, p_out_a(X)) -> p_out_a(X) The argument filtering Pi contains the following mapping: p_in_a(x1) = p_in_a U1_a(x1, x2) = U1_a(x2) q_in_g(x1) = q_in_g(x1) b = b q_out_g(x1) = q_out_g a = a p_out_a(x1) = p_out_a U2_a(x1, x2) = U2_a(x2) P_IN_A(x1) = P_IN_A We have to consider all (P,R,Pi)-chains ---------------------------------------- (9) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (10) Obligation: Pi DP problem: The TRS P consists of the following rules: P_IN_A(X) -> P_IN_A(X) R is empty. The argument filtering Pi contains the following mapping: P_IN_A(x1) = P_IN_A We have to consider all (P,R,Pi)-chains ---------------------------------------- (11) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: P_IN_A -> P_IN_A R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (13) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by semiunifying a rule from P directly. s = P_IN_A evaluates to t =P_IN_A Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence The DP semiunifies directly so there is only one rewrite step from P_IN_A to P_IN_A. ---------------------------------------- (14) NO ---------------------------------------- (15) PrologToPiTRSProof (SOUND) We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: p_in_1: (f) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: p_in_a(X) -> U1_a(X, q_in_g(a)) q_in_g(b) -> q_out_g(b) U1_a(X, q_out_g(a)) -> p_out_a(X) p_in_a(X) -> U2_a(X, p_in_a(X)) U2_a(X, p_out_a(X)) -> p_out_a(X) The argument filtering Pi contains the following mapping: p_in_a(x1) = p_in_a U1_a(x1, x2) = U1_a(x2) q_in_g(x1) = q_in_g(x1) b = b q_out_g(x1) = q_out_g(x1) a = a p_out_a(x1) = p_out_a U2_a(x1, x2) = U2_a(x2) Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (16) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: p_in_a(X) -> U1_a(X, q_in_g(a)) q_in_g(b) -> q_out_g(b) U1_a(X, q_out_g(a)) -> p_out_a(X) p_in_a(X) -> U2_a(X, p_in_a(X)) U2_a(X, p_out_a(X)) -> p_out_a(X) The argument filtering Pi contains the following mapping: p_in_a(x1) = p_in_a U1_a(x1, x2) = U1_a(x2) q_in_g(x1) = q_in_g(x1) b = b q_out_g(x1) = q_out_g(x1) a = a p_out_a(x1) = p_out_a U2_a(x1, x2) = U2_a(x2) ---------------------------------------- (17) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem: Pi DP problem: The TRS P consists of the following rules: P_IN_A(X) -> U1_A(X, q_in_g(a)) P_IN_A(X) -> Q_IN_G(a) P_IN_A(X) -> U2_A(X, p_in_a(X)) P_IN_A(X) -> P_IN_A(X) The TRS R consists of the following rules: p_in_a(X) -> U1_a(X, q_in_g(a)) q_in_g(b) -> q_out_g(b) U1_a(X, q_out_g(a)) -> p_out_a(X) p_in_a(X) -> U2_a(X, p_in_a(X)) U2_a(X, p_out_a(X)) -> p_out_a(X) The argument filtering Pi contains the following mapping: p_in_a(x1) = p_in_a U1_a(x1, x2) = U1_a(x2) q_in_g(x1) = q_in_g(x1) b = b q_out_g(x1) = q_out_g(x1) a = a p_out_a(x1) = p_out_a U2_a(x1, x2) = U2_a(x2) P_IN_A(x1) = P_IN_A U1_A(x1, x2) = U1_A(x2) Q_IN_G(x1) = Q_IN_G(x1) U2_A(x1, x2) = U2_A(x2) We have to consider all (P,R,Pi)-chains ---------------------------------------- (18) Obligation: Pi DP problem: The TRS P consists of the following rules: P_IN_A(X) -> U1_A(X, q_in_g(a)) P_IN_A(X) -> Q_IN_G(a) P_IN_A(X) -> U2_A(X, p_in_a(X)) P_IN_A(X) -> P_IN_A(X) The TRS R consists of the following rules: p_in_a(X) -> U1_a(X, q_in_g(a)) q_in_g(b) -> q_out_g(b) U1_a(X, q_out_g(a)) -> p_out_a(X) p_in_a(X) -> U2_a(X, p_in_a(X)) U2_a(X, p_out_a(X)) -> p_out_a(X) The argument filtering Pi contains the following mapping: p_in_a(x1) = p_in_a U1_a(x1, x2) = U1_a(x2) q_in_g(x1) = q_in_g(x1) b = b q_out_g(x1) = q_out_g(x1) a = a p_out_a(x1) = p_out_a U2_a(x1, x2) = U2_a(x2) P_IN_A(x1) = P_IN_A U1_A(x1, x2) = U1_A(x2) Q_IN_G(x1) = Q_IN_G(x1) U2_A(x1, x2) = U2_A(x2) We have to consider all (P,R,Pi)-chains ---------------------------------------- (19) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes. ---------------------------------------- (20) Obligation: Pi DP problem: The TRS P consists of the following rules: P_IN_A(X) -> P_IN_A(X) The TRS R consists of the following rules: p_in_a(X) -> U1_a(X, q_in_g(a)) q_in_g(b) -> q_out_g(b) U1_a(X, q_out_g(a)) -> p_out_a(X) p_in_a(X) -> U2_a(X, p_in_a(X)) U2_a(X, p_out_a(X)) -> p_out_a(X) The argument filtering Pi contains the following mapping: p_in_a(x1) = p_in_a U1_a(x1, x2) = U1_a(x2) q_in_g(x1) = q_in_g(x1) b = b q_out_g(x1) = q_out_g(x1) a = a p_out_a(x1) = p_out_a U2_a(x1, x2) = U2_a(x2) P_IN_A(x1) = P_IN_A We have to consider all (P,R,Pi)-chains ---------------------------------------- (21) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (22) Obligation: Pi DP problem: The TRS P consists of the following rules: P_IN_A(X) -> P_IN_A(X) R is empty. The argument filtering Pi contains the following mapping: P_IN_A(x1) = P_IN_A We have to consider all (P,R,Pi)-chains ---------------------------------------- (23) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (24) Obligation: Q DP problem: The TRS P consists of the following rules: P_IN_A -> P_IN_A R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (25) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by semiunifying a rule from P directly. s = P_IN_A evaluates to t =P_IN_A Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence The DP semiunifies directly so there is only one rewrite step from P_IN_A to P_IN_A. ---------------------------------------- (26) NO ---------------------------------------- (27) PrologToTRSTransformerProof (SOUND) Transformed Prolog program to TRS. { "root": 3, "program": { "directives": [], "clauses": [ [ "(p X)", "(',' (q (a)) (!))" ], [ "(p X)", "(p X)" ], [ "(q (b))", null ] ] }, "graph": { "nodes": { "3": { "goal": [{ "clause": -1, "scope": -1, "term": "(p T1)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "27": { "goal": [ { "clause": 2, "scope": 2, "term": "(',' (q (a)) (!_1))" }, { "clause": -1, "scope": 2, "term": null }, { "clause": 1, "scope": 1, "term": "(p T1)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "6": { "goal": [ { "clause": 0, "scope": 1, "term": "(p T1)" }, { "clause": 1, "scope": 1, "term": "(p T1)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "28": { "goal": [ { "clause": -1, "scope": 2, "term": null }, { "clause": 1, "scope": 1, "term": "(p T1)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "29": { "goal": [{ "clause": 1, "scope": 1, "term": "(p T1)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "type": "Nodes", "30": { "goal": [{ "clause": -1, "scope": -1, "term": "(p T10)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "20": { "goal": [ { "clause": -1, "scope": -1, "term": "(',' (q (a)) (!_1))" }, { "clause": 1, "scope": 1, "term": "(p T1)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } } }, "edges": [ { "from": 3, "to": 6, "label": "CASE" }, { "from": 6, "to": 20, "label": "ONLY EVAL with clause\np(X3) :- ','(q(a), !_1).\nand substitutionT1 -> T4,\nX3 -> T4" }, { "from": 20, "to": 27, "label": "CASE" }, { "from": 27, "to": 28, "label": "BACKTRACK\nfor clause: q(b)because of non-unification" }, { "from": 28, "to": 29, "label": "FAILURE" }, { "from": 29, "to": 30, "label": "ONLY EVAL with clause\np(X7) :- p(X7).\nand substitutionT1 -> T10,\nX7 -> T10,\nT9 -> T10" }, { "from": 30, "to": 3, "label": "INSTANCE with matching:\nT1 -> T10" } ], "type": "Graph" } } ---------------------------------------- (28) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f3_in -> U1(f3_in) U1(f3_out1) -> f3_out1 Q is empty. ---------------------------------------- (29) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(U1(x_1)) = 2*x_1 POL(f3_in) = 0 POL(f3_out1) = 1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: U1(f3_out1) -> f3_out1 ---------------------------------------- (30) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f3_in -> U1(f3_in) Q is empty. ---------------------------------------- (31) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (32) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f3_in -> U1(f3_in) The set Q consists of the following terms: f3_in ---------------------------------------- (33) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (34) Obligation: Q DP problem: The TRS P consists of the following rules: F3_IN -> F3_IN The TRS R consists of the following rules: f3_in -> U1(f3_in) The set Q consists of the following terms: f3_in We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (35) 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. ---------------------------------------- (36) Obligation: Q DP problem: The TRS P consists of the following rules: F3_IN -> F3_IN R is empty. The set Q consists of the following terms: f3_in We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (37) 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]. f3_in ---------------------------------------- (38) Obligation: Q DP problem: The TRS P consists of the following rules: F3_IN -> F3_IN R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (39) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by semiunifying a rule from P directly. s = F3_IN evaluates to t =F3_IN Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence The DP semiunifies directly so there is only one rewrite step from F3_IN to F3_IN. ---------------------------------------- (40) NO ---------------------------------------- (41) PrologToDTProblemTransformerProof (SOUND) Built DT problem from termination graph DT10. { "root": 1, "program": { "directives": [], "clauses": [ [ "(p X)", "(',' (q (a)) (!))" ], [ "(p X)", "(p X)" ], [ "(q (b))", null ] ] }, "graph": { "nodes": { "1": { "goal": [{ "clause": -1, "scope": -1, "term": "(p T1)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "23": { "goal": [ { "clause": -1, "scope": 2, "term": null }, { "clause": 1, "scope": 1, "term": "(p T1)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "26": { "goal": [{ "clause": 1, "scope": 1, "term": "(p T1)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "5": { "goal": [ { "clause": 0, "scope": 1, "term": "(p T1)" }, { "clause": 1, "scope": 1, "term": "(p T1)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "19": { "goal": [ { "clause": -1, "scope": -1, "term": "(',' (q (a)) (!_1))" }, { "clause": 1, "scope": 1, "term": "(p T1)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "type": "Nodes", "31": { "goal": [{ "clause": -1, "scope": -1, "term": "(p T9)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "21": { "goal": [ { "clause": 2, "scope": 2, "term": "(',' (q (a)) (!_1))" }, { "clause": -1, "scope": 2, "term": null }, { "clause": 1, "scope": 1, "term": "(p T1)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } } }, "edges": [ { "from": 1, "to": 5, "label": "CASE" }, { "from": 5, "to": 19, "label": "ONLY EVAL with clause\np(X2) :- ','(q(a), !_1).\nand substitutionT1 -> T3,\nX2 -> T3" }, { "from": 19, "to": 21, "label": "CASE" }, { "from": 21, "to": 23, "label": "BACKTRACK\nfor clause: q(b)because of non-unification" }, { "from": 23, "to": 26, "label": "FAILURE" }, { "from": 26, "to": 31, "label": "ONLY EVAL with clause\np(X6) :- p(X6).\nand substitutionT1 -> T9,\nX6 -> T9,\nT8 -> T9" }, { "from": 31, "to": 1, "label": "INSTANCE with matching:\nT1 -> T9" } ], "type": "Graph" } } ---------------------------------------- (42) Obligation: Triples: pA(X1) :- pA(X1). Clauses: pcA(X1) :- pcA(X1). Afs: pA(x1) = pA ---------------------------------------- (43) 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: (f) Transforming TRIPLES into the following Term Rewriting System: Pi DP problem: The TRS P consists of the following rules: PA_IN_A(X1) -> U1_A(X1, pA_in_a(X1)) PA_IN_A(X1) -> PA_IN_A(X1) R is empty. The argument filtering Pi contains the following mapping: pA_in_a(x1) = pA_in_a PA_IN_A(x1) = PA_IN_A U1_A(x1, x2) = U1_A(x2) We have to consider all (P,R,Pi)-chains Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES ---------------------------------------- (44) Obligation: Pi DP problem: The TRS P consists of the following rules: PA_IN_A(X1) -> U1_A(X1, pA_in_a(X1)) PA_IN_A(X1) -> PA_IN_A(X1) R is empty. The argument filtering Pi contains the following mapping: pA_in_a(x1) = pA_in_a PA_IN_A(x1) = PA_IN_A U1_A(x1, x2) = U1_A(x2) We have to consider all (P,R,Pi)-chains ---------------------------------------- (45) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node. ---------------------------------------- (46) Obligation: Pi DP problem: The TRS P consists of the following rules: PA_IN_A(X1) -> PA_IN_A(X1) R is empty. The argument filtering Pi contains the following mapping: PA_IN_A(x1) = PA_IN_A We have to consider all (P,R,Pi)-chains ---------------------------------------- (47) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (48) Obligation: Q DP problem: The TRS P consists of the following rules: PA_IN_A -> PA_IN_A R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (49) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by semiunifying a rule from P directly. s = PA_IN_A evaluates to t =PA_IN_A Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence The DP semiunifies directly so there is only one rewrite step from PA_IN_A to PA_IN_A. ---------------------------------------- (50) NO