/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.pl /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- MAYBE proof of /export/starexec/sandbox2/benchmark/theBenchmark.pl # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Left Termination of the query pattern p(a,a,a) w.r.t. the given Prolog program could not be shown: (0) Prolog (1) PrologToPiTRSProof [SOUND, 0 ms] (2) PiTRS (3) DependencyPairsProof [EQUIVALENT, 9 ms] (4) PiDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) PiDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) PiDP (9) PiDPToQDPProof [SOUND, 0 ms] (10) QDP (11) PrologToTRSTransformerProof [SOUND, 0 ms] (12) QTRS (13) QTRSRRRProof [EQUIVALENT, 43 ms] (14) QTRS (15) Overlay + Local Confluence [EQUIVALENT, 0 ms] (16) QTRS (17) DependencyPairsProof [EQUIVALENT, 0 ms] (18) QDP (19) UsableRulesProof [EQUIVALENT, 0 ms] (20) QDP (21) QReductionProof [EQUIVALENT, 0 ms] (22) QDP (23) PrologToPiTRSProof [SOUND, 0 ms] (24) PiTRS (25) DependencyPairsProof [EQUIVALENT, 4 ms] (26) PiDP (27) DependencyGraphProof [EQUIVALENT, 0 ms] (28) PiDP (29) UsableRulesProof [EQUIVALENT, 0 ms] (30) PiDP (31) PiDPToQDPProof [SOUND, 0 ms] (32) QDP (33) PrologToDTProblemTransformerProof [SOUND, 0 ms] (34) TRIPLES (35) TriplesToPiDPProof [SOUND, 5 ms] (36) PiDP (37) DependencyGraphProof [EQUIVALENT, 0 ms] (38) PiDP (39) PiDPToQDPProof [SOUND, 0 ms] (40) QDP (41) PrologToIRSwTTransformerProof [SOUND, 0 ms] (42) IRSwT (43) IRSwTSimpleDependencyGraphProof [EQUIVALENT, 0 ms] (44) IRSwT (45) IntTRSCompressionProof [EQUIVALENT, 25 ms] (46) IRSwT (47) IRSFormatTransformerProof [EQUIVALENT, 0 ms] (48) IRSwT (49) IRSwTTerminationDigraphProof [EQUIVALENT, 0 ms] (50) IRSwT (51) FilterProof [EQUIVALENT, 1 ms] (52) IntTRS (53) IntTRSNonPeriodicNontermProof [COMPLETE, 0 ms] (54) NO ---------------------------------------- (0) Obligation: Clauses: p(X, Y, Z) :- append(.(X, Y), Z, Y). append([], Y, Y). append(.(H, Xs), Ys, .(H, Zs)) :- append(Xs, Ys, Zs). Query: p(a,a,a) ---------------------------------------- (1) 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_3: (f,f,f) append_in_3: (f,f,f) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: p_in_aaa(X, Y, Z) -> U1_aaa(X, Y, Z, append_in_aaa(.(X, Y), Z, Y)) append_in_aaa([], Y, Y) -> append_out_aaa([], Y, Y) append_in_aaa(.(H, Xs), Ys, .(H, Zs)) -> U2_aaa(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs)) U2_aaa(H, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) -> append_out_aaa(.(H, Xs), Ys, .(H, Zs)) U1_aaa(X, Y, Z, append_out_aaa(.(X, Y), Z, Y)) -> p_out_aaa(X, Y, Z) The argument filtering Pi contains the following mapping: p_in_aaa(x1, x2, x3) = p_in_aaa U1_aaa(x1, x2, x3, x4) = U1_aaa(x4) append_in_aaa(x1, x2, x3) = append_in_aaa .(x1, x2) = .(x2) append_out_aaa(x1, x2, x3) = append_out_aaa(x1) U2_aaa(x1, x2, x3, x4, x5) = U2_aaa(x5) p_out_aaa(x1, x2, x3) = p_out_aaa(x2) Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (2) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: p_in_aaa(X, Y, Z) -> U1_aaa(X, Y, Z, append_in_aaa(.(X, Y), Z, Y)) append_in_aaa([], Y, Y) -> append_out_aaa([], Y, Y) append_in_aaa(.(H, Xs), Ys, .(H, Zs)) -> U2_aaa(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs)) U2_aaa(H, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) -> append_out_aaa(.(H, Xs), Ys, .(H, Zs)) U1_aaa(X, Y, Z, append_out_aaa(.(X, Y), Z, Y)) -> p_out_aaa(X, Y, Z) The argument filtering Pi contains the following mapping: p_in_aaa(x1, x2, x3) = p_in_aaa U1_aaa(x1, x2, x3, x4) = U1_aaa(x4) append_in_aaa(x1, x2, x3) = append_in_aaa .(x1, x2) = .(x2) append_out_aaa(x1, x2, x3) = append_out_aaa(x1) U2_aaa(x1, x2, x3, x4, x5) = U2_aaa(x5) p_out_aaa(x1, x2, x3) = p_out_aaa(x2) ---------------------------------------- (3) 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_AAA(X, Y, Z) -> U1_AAA(X, Y, Z, append_in_aaa(.(X, Y), Z, Y)) P_IN_AAA(X, Y, Z) -> APPEND_IN_AAA(.(X, Y), Z, Y) APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) -> U2_AAA(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs)) APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) -> APPEND_IN_AAA(Xs, Ys, Zs) The TRS R consists of the following rules: p_in_aaa(X, Y, Z) -> U1_aaa(X, Y, Z, append_in_aaa(.(X, Y), Z, Y)) append_in_aaa([], Y, Y) -> append_out_aaa([], Y, Y) append_in_aaa(.(H, Xs), Ys, .(H, Zs)) -> U2_aaa(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs)) U2_aaa(H, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) -> append_out_aaa(.(H, Xs), Ys, .(H, Zs)) U1_aaa(X, Y, Z, append_out_aaa(.(X, Y), Z, Y)) -> p_out_aaa(X, Y, Z) The argument filtering Pi contains the following mapping: p_in_aaa(x1, x2, x3) = p_in_aaa U1_aaa(x1, x2, x3, x4) = U1_aaa(x4) append_in_aaa(x1, x2, x3) = append_in_aaa .(x1, x2) = .(x2) append_out_aaa(x1, x2, x3) = append_out_aaa(x1) U2_aaa(x1, x2, x3, x4, x5) = U2_aaa(x5) p_out_aaa(x1, x2, x3) = p_out_aaa(x2) P_IN_AAA(x1, x2, x3) = P_IN_AAA U1_AAA(x1, x2, x3, x4) = U1_AAA(x4) APPEND_IN_AAA(x1, x2, x3) = APPEND_IN_AAA U2_AAA(x1, x2, x3, x4, x5) = U2_AAA(x5) We have to consider all (P,R,Pi)-chains ---------------------------------------- (4) Obligation: Pi DP problem: The TRS P consists of the following rules: P_IN_AAA(X, Y, Z) -> U1_AAA(X, Y, Z, append_in_aaa(.(X, Y), Z, Y)) P_IN_AAA(X, Y, Z) -> APPEND_IN_AAA(.(X, Y), Z, Y) APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) -> U2_AAA(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs)) APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) -> APPEND_IN_AAA(Xs, Ys, Zs) The TRS R consists of the following rules: p_in_aaa(X, Y, Z) -> U1_aaa(X, Y, Z, append_in_aaa(.(X, Y), Z, Y)) append_in_aaa([], Y, Y) -> append_out_aaa([], Y, Y) append_in_aaa(.(H, Xs), Ys, .(H, Zs)) -> U2_aaa(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs)) U2_aaa(H, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) -> append_out_aaa(.(H, Xs), Ys, .(H, Zs)) U1_aaa(X, Y, Z, append_out_aaa(.(X, Y), Z, Y)) -> p_out_aaa(X, Y, Z) The argument filtering Pi contains the following mapping: p_in_aaa(x1, x2, x3) = p_in_aaa U1_aaa(x1, x2, x3, x4) = U1_aaa(x4) append_in_aaa(x1, x2, x3) = append_in_aaa .(x1, x2) = .(x2) append_out_aaa(x1, x2, x3) = append_out_aaa(x1) U2_aaa(x1, x2, x3, x4, x5) = U2_aaa(x5) p_out_aaa(x1, x2, x3) = p_out_aaa(x2) P_IN_AAA(x1, x2, x3) = P_IN_AAA U1_AAA(x1, x2, x3, x4) = U1_AAA(x4) APPEND_IN_AAA(x1, x2, x3) = APPEND_IN_AAA U2_AAA(x1, x2, x3, x4, x5) = U2_AAA(x5) We have to consider all (P,R,Pi)-chains ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes. ---------------------------------------- (6) Obligation: Pi DP problem: The TRS P consists of the following rules: APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) -> APPEND_IN_AAA(Xs, Ys, Zs) The TRS R consists of the following rules: p_in_aaa(X, Y, Z) -> U1_aaa(X, Y, Z, append_in_aaa(.(X, Y), Z, Y)) append_in_aaa([], Y, Y) -> append_out_aaa([], Y, Y) append_in_aaa(.(H, Xs), Ys, .(H, Zs)) -> U2_aaa(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs)) U2_aaa(H, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) -> append_out_aaa(.(H, Xs), Ys, .(H, Zs)) U1_aaa(X, Y, Z, append_out_aaa(.(X, Y), Z, Y)) -> p_out_aaa(X, Y, Z) The argument filtering Pi contains the following mapping: p_in_aaa(x1, x2, x3) = p_in_aaa U1_aaa(x1, x2, x3, x4) = U1_aaa(x4) append_in_aaa(x1, x2, x3) = append_in_aaa .(x1, x2) = .(x2) append_out_aaa(x1, x2, x3) = append_out_aaa(x1) U2_aaa(x1, x2, x3, x4, x5) = U2_aaa(x5) p_out_aaa(x1, x2, x3) = p_out_aaa(x2) APPEND_IN_AAA(x1, x2, x3) = APPEND_IN_AAA We have to consider all (P,R,Pi)-chains ---------------------------------------- (7) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (8) Obligation: Pi DP problem: The TRS P consists of the following rules: APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) -> APPEND_IN_AAA(Xs, Ys, Zs) R is empty. The argument filtering Pi contains the following mapping: .(x1, x2) = .(x2) APPEND_IN_AAA(x1, x2, x3) = APPEND_IN_AAA We have to consider all (P,R,Pi)-chains ---------------------------------------- (9) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: APPEND_IN_AAA -> APPEND_IN_AAA R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (11) PrologToTRSTransformerProof (SOUND) Transformed Prolog program to TRS. { "root": 1, "program": { "directives": [], "clauses": [ [ "(p X Y Z)", "(append (. X Y) Z Y)" ], [ "(append ([]) Y Y)", null ], [ "(append (. H Xs) Ys (. H Zs))", "(append Xs Ys Zs)" ] ] }, "graph": { "nodes": { "121": { "goal": [{ "clause": 2, "scope": 2, "term": "(append (. T18 T17) T16 T17)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "1": { "goal": [{ "clause": -1, "scope": -1, "term": "(p T1 T2 T3)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "12": { "goal": [{ "clause": 0, "scope": 1, "term": "(p T1 T2 T3)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "122": { "goal": [{ "clause": -1, "scope": -1, "term": "(append (. T33 T32) T31 T32)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "123": { "goal": [], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "118": { "goal": [{ "clause": -1, "scope": -1, "term": "(append (. T18 T17) T16 T17)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "type": "Nodes", "120": { "goal": [ { "clause": 1, "scope": 2, "term": "(append (. T18 T17) T16 T17)" }, { "clause": 2, "scope": 2, "term": "(append (. T18 T17) T16 T17)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } } }, "edges": [ { "from": 1, "to": 12, "label": "CASE" }, { "from": 12, "to": 118, "label": "ONLY EVAL with clause\np(X9, X10, X11) :- append(.(X9, X10), X11, X10).\nand substitutionT1 -> T18,\nX9 -> T18,\nT2 -> T17,\nX10 -> T17,\nT3 -> T16,\nX11 -> T16,\nT15 -> T16,\nT14 -> T17,\nT13 -> T18" }, { "from": 118, "to": 120, "label": "CASE" }, { "from": 120, "to": 121, "label": "BACKTRACK\nfor clause: append([], Y, Y)because of non-unification" }, { "from": 121, "to": 122, "label": "EVAL with clause\nappend(.(X23, X24), X25, .(X23, X26)) :- append(X24, X25, X26).\nand substitutionT18 -> T33,\nX23 -> T33,\nT17 -> .(T33, T32),\nX24 -> .(T33, T32),\nT16 -> T31,\nX25 -> T31,\nX26 -> T32,\nT28 -> .(T33, T32),\nT29 -> T31,\nT30 -> T32,\nT27 -> T33" }, { "from": 121, "to": 123, "label": "EVAL-BACKTRACK" }, { "from": 122, "to": 118, "label": "INSTANCE with matching:\nT18 -> T33\nT17 -> T32\nT16 -> T31" } ], "type": "Graph" } } ---------------------------------------- (12) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f1_in -> U1(f118_in) U1(f118_out1) -> f1_out1 f118_in -> U2(f118_in) U2(f118_out1) -> f118_out1 Q is empty. ---------------------------------------- (13) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(U1(x_1)) = 2*x_1 POL(U2(x_1)) = 2*x_1 POL(f118_in) = 0 POL(f118_out1) = 2 POL(f1_in) = 1 POL(f1_out1) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: f1_in -> U1(f118_in) U1(f118_out1) -> f1_out1 U2(f118_out1) -> f118_out1 ---------------------------------------- (14) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f118_in -> U2(f118_in) Q is empty. ---------------------------------------- (15) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (16) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f118_in -> U2(f118_in) The set Q consists of the following terms: f118_in ---------------------------------------- (17) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: F118_IN -> F118_IN The TRS R consists of the following rules: f118_in -> U2(f118_in) The set Q consists of the following terms: f118_in We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) 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. ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: F118_IN -> F118_IN R is empty. The set Q consists of the following terms: f118_in We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) 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]. f118_in ---------------------------------------- (22) Obligation: Q DP problem: The TRS P consists of the following rules: F118_IN -> F118_IN R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (23) 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_3: (f,f,f) append_in_3: (f,f,f) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: p_in_aaa(X, Y, Z) -> U1_aaa(X, Y, Z, append_in_aaa(.(X, Y), Z, Y)) append_in_aaa([], Y, Y) -> append_out_aaa([], Y, Y) append_in_aaa(.(H, Xs), Ys, .(H, Zs)) -> U2_aaa(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs)) U2_aaa(H, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) -> append_out_aaa(.(H, Xs), Ys, .(H, Zs)) U1_aaa(X, Y, Z, append_out_aaa(.(X, Y), Z, Y)) -> p_out_aaa(X, Y, Z) The argument filtering Pi contains the following mapping: p_in_aaa(x1, x2, x3) = p_in_aaa U1_aaa(x1, x2, x3, x4) = U1_aaa(x4) append_in_aaa(x1, x2, x3) = append_in_aaa .(x1, x2) = .(x2) append_out_aaa(x1, x2, x3) = append_out_aaa(x1) U2_aaa(x1, x2, x3, x4, x5) = U2_aaa(x5) p_out_aaa(x1, x2, x3) = p_out_aaa(x2) Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (24) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: p_in_aaa(X, Y, Z) -> U1_aaa(X, Y, Z, append_in_aaa(.(X, Y), Z, Y)) append_in_aaa([], Y, Y) -> append_out_aaa([], Y, Y) append_in_aaa(.(H, Xs), Ys, .(H, Zs)) -> U2_aaa(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs)) U2_aaa(H, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) -> append_out_aaa(.(H, Xs), Ys, .(H, Zs)) U1_aaa(X, Y, Z, append_out_aaa(.(X, Y), Z, Y)) -> p_out_aaa(X, Y, Z) The argument filtering Pi contains the following mapping: p_in_aaa(x1, x2, x3) = p_in_aaa U1_aaa(x1, x2, x3, x4) = U1_aaa(x4) append_in_aaa(x1, x2, x3) = append_in_aaa .(x1, x2) = .(x2) append_out_aaa(x1, x2, x3) = append_out_aaa(x1) U2_aaa(x1, x2, x3, x4, x5) = U2_aaa(x5) p_out_aaa(x1, x2, x3) = p_out_aaa(x2) ---------------------------------------- (25) 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_AAA(X, Y, Z) -> U1_AAA(X, Y, Z, append_in_aaa(.(X, Y), Z, Y)) P_IN_AAA(X, Y, Z) -> APPEND_IN_AAA(.(X, Y), Z, Y) APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) -> U2_AAA(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs)) APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) -> APPEND_IN_AAA(Xs, Ys, Zs) The TRS R consists of the following rules: p_in_aaa(X, Y, Z) -> U1_aaa(X, Y, Z, append_in_aaa(.(X, Y), Z, Y)) append_in_aaa([], Y, Y) -> append_out_aaa([], Y, Y) append_in_aaa(.(H, Xs), Ys, .(H, Zs)) -> U2_aaa(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs)) U2_aaa(H, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) -> append_out_aaa(.(H, Xs), Ys, .(H, Zs)) U1_aaa(X, Y, Z, append_out_aaa(.(X, Y), Z, Y)) -> p_out_aaa(X, Y, Z) The argument filtering Pi contains the following mapping: p_in_aaa(x1, x2, x3) = p_in_aaa U1_aaa(x1, x2, x3, x4) = U1_aaa(x4) append_in_aaa(x1, x2, x3) = append_in_aaa .(x1, x2) = .(x2) append_out_aaa(x1, x2, x3) = append_out_aaa(x1) U2_aaa(x1, x2, x3, x4, x5) = U2_aaa(x5) p_out_aaa(x1, x2, x3) = p_out_aaa(x2) P_IN_AAA(x1, x2, x3) = P_IN_AAA U1_AAA(x1, x2, x3, x4) = U1_AAA(x4) APPEND_IN_AAA(x1, x2, x3) = APPEND_IN_AAA U2_AAA(x1, x2, x3, x4, x5) = U2_AAA(x5) We have to consider all (P,R,Pi)-chains ---------------------------------------- (26) Obligation: Pi DP problem: The TRS P consists of the following rules: P_IN_AAA(X, Y, Z) -> U1_AAA(X, Y, Z, append_in_aaa(.(X, Y), Z, Y)) P_IN_AAA(X, Y, Z) -> APPEND_IN_AAA(.(X, Y), Z, Y) APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) -> U2_AAA(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs)) APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) -> APPEND_IN_AAA(Xs, Ys, Zs) The TRS R consists of the following rules: p_in_aaa(X, Y, Z) -> U1_aaa(X, Y, Z, append_in_aaa(.(X, Y), Z, Y)) append_in_aaa([], Y, Y) -> append_out_aaa([], Y, Y) append_in_aaa(.(H, Xs), Ys, .(H, Zs)) -> U2_aaa(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs)) U2_aaa(H, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) -> append_out_aaa(.(H, Xs), Ys, .(H, Zs)) U1_aaa(X, Y, Z, append_out_aaa(.(X, Y), Z, Y)) -> p_out_aaa(X, Y, Z) The argument filtering Pi contains the following mapping: p_in_aaa(x1, x2, x3) = p_in_aaa U1_aaa(x1, x2, x3, x4) = U1_aaa(x4) append_in_aaa(x1, x2, x3) = append_in_aaa .(x1, x2) = .(x2) append_out_aaa(x1, x2, x3) = append_out_aaa(x1) U2_aaa(x1, x2, x3, x4, x5) = U2_aaa(x5) p_out_aaa(x1, x2, x3) = p_out_aaa(x2) P_IN_AAA(x1, x2, x3) = P_IN_AAA U1_AAA(x1, x2, x3, x4) = U1_AAA(x4) APPEND_IN_AAA(x1, x2, x3) = APPEND_IN_AAA U2_AAA(x1, x2, x3, x4, x5) = U2_AAA(x5) We have to consider all (P,R,Pi)-chains ---------------------------------------- (27) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes. ---------------------------------------- (28) Obligation: Pi DP problem: The TRS P consists of the following rules: APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) -> APPEND_IN_AAA(Xs, Ys, Zs) The TRS R consists of the following rules: p_in_aaa(X, Y, Z) -> U1_aaa(X, Y, Z, append_in_aaa(.(X, Y), Z, Y)) append_in_aaa([], Y, Y) -> append_out_aaa([], Y, Y) append_in_aaa(.(H, Xs), Ys, .(H, Zs)) -> U2_aaa(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs)) U2_aaa(H, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) -> append_out_aaa(.(H, Xs), Ys, .(H, Zs)) U1_aaa(X, Y, Z, append_out_aaa(.(X, Y), Z, Y)) -> p_out_aaa(X, Y, Z) The argument filtering Pi contains the following mapping: p_in_aaa(x1, x2, x3) = p_in_aaa U1_aaa(x1, x2, x3, x4) = U1_aaa(x4) append_in_aaa(x1, x2, x3) = append_in_aaa .(x1, x2) = .(x2) append_out_aaa(x1, x2, x3) = append_out_aaa(x1) U2_aaa(x1, x2, x3, x4, x5) = U2_aaa(x5) p_out_aaa(x1, x2, x3) = p_out_aaa(x2) APPEND_IN_AAA(x1, x2, x3) = APPEND_IN_AAA We have to consider all (P,R,Pi)-chains ---------------------------------------- (29) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (30) Obligation: Pi DP problem: The TRS P consists of the following rules: APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) -> APPEND_IN_AAA(Xs, Ys, Zs) R is empty. The argument filtering Pi contains the following mapping: .(x1, x2) = .(x2) APPEND_IN_AAA(x1, x2, x3) = APPEND_IN_AAA We have to consider all (P,R,Pi)-chains ---------------------------------------- (31) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (32) Obligation: Q DP problem: The TRS P consists of the following rules: APPEND_IN_AAA -> APPEND_IN_AAA R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (33) PrologToDTProblemTransformerProof (SOUND) Built DT problem from termination graph DT10. { "root": 2, "program": { "directives": [], "clauses": [ [ "(p X Y Z)", "(append (. X Y) Z Y)" ], [ "(append ([]) Y Y)", null ], [ "(append (. H Xs) Ys (. H Zs))", "(append Xs Ys Zs)" ] ] }, "graph": { "nodes": { "2": { "goal": [{ "clause": -1, "scope": -1, "term": "(p T1 T2 T3)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "103": { "goal": [{ "clause": -1, "scope": -1, "term": "(append (. T12 T11) T10 T11)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "5": { "goal": [{ "clause": 0, "scope": 1, "term": "(p T1 T2 T3)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "105": { "goal": [ { "clause": 1, "scope": 2, "term": "(append (. T12 T11) T10 T11)" }, { "clause": 2, "scope": 2, "term": "(append (. T12 T11) T10 T11)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "116": { "goal": [{ "clause": -1, "scope": -1, "term": "(append (. T27 T26) T25 T26)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "106": { "goal": [{ "clause": 2, "scope": 2, "term": "(append (. T12 T11) T10 T11)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "117": { "goal": [], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "type": "Nodes" }, "edges": [ { "from": 2, "to": 5, "label": "CASE" }, { "from": 5, "to": 103, "label": "ONLY EVAL with clause\np(X4, X5, X6) :- append(.(X4, X5), X6, X5).\nand substitutionT1 -> T12,\nX4 -> T12,\nT2 -> T11,\nX5 -> T11,\nT3 -> T10,\nX6 -> T10,\nT9 -> T10,\nT8 -> T11,\nT7 -> T12" }, { "from": 103, "to": 105, "label": "CASE" }, { "from": 105, "to": 106, "label": "BACKTRACK\nfor clause: append([], Y, Y)because of non-unification" }, { "from": 106, "to": 116, "label": "EVAL with clause\nappend(.(X16, X17), X18, .(X16, X19)) :- append(X17, X18, X19).\nand substitutionT12 -> T27,\nX16 -> T27,\nT11 -> .(T27, T26),\nX17 -> .(T27, T26),\nT10 -> T25,\nX18 -> T25,\nX19 -> T26,\nT22 -> .(T27, T26),\nT23 -> T25,\nT24 -> T26,\nT21 -> T27" }, { "from": 106, "to": 117, "label": "EVAL-BACKTRACK" }, { "from": 116, "to": 103, "label": "INSTANCE with matching:\nT12 -> T27\nT11 -> T26\nT10 -> T25" } ], "type": "Graph" } } ---------------------------------------- (34) Obligation: Triples: appendA(X1, .(X1, X2), X3) :- appendA(X1, X2, X3). pB(X1, X2, X3) :- appendA(X1, X2, X3). Clauses: appendcA(X1, .(X1, X2), X3) :- appendcA(X1, X2, X3). Afs: pB(x1, x2, x3) = pB ---------------------------------------- (35) TriplesToPiDPProof (SOUND) We use the technique of [DT09]. With regard to the inferred argument filtering the predicates were used in the following modes: pB_in_3: (f,f,f) appendA_in_3: (f,f,f) Transforming TRIPLES into the following Term Rewriting System: Pi DP problem: The TRS P consists of the following rules: PB_IN_AAA(X1, X2, X3) -> U2_AAA(X1, X2, X3, appendA_in_aaa(X1, X2, X3)) PB_IN_AAA(X1, X2, X3) -> APPENDA_IN_AAA(X1, X2, X3) APPENDA_IN_AAA(X1, .(X1, X2), X3) -> U1_AAA(X1, X2, X3, appendA_in_aaa(X1, X2, X3)) APPENDA_IN_AAA(X1, .(X1, X2), X3) -> APPENDA_IN_AAA(X1, X2, X3) R is empty. The argument filtering Pi contains the following mapping: appendA_in_aaa(x1, x2, x3) = appendA_in_aaa .(x1, x2) = .(x1, x2) PB_IN_AAA(x1, x2, x3) = PB_IN_AAA U2_AAA(x1, x2, x3, x4) = U2_AAA(x4) APPENDA_IN_AAA(x1, x2, x3) = APPENDA_IN_AAA U1_AAA(x1, x2, x3, x4) = U1_AAA(x4) We have to consider all (P,R,Pi)-chains Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES ---------------------------------------- (36) Obligation: Pi DP problem: The TRS P consists of the following rules: PB_IN_AAA(X1, X2, X3) -> U2_AAA(X1, X2, X3, appendA_in_aaa(X1, X2, X3)) PB_IN_AAA(X1, X2, X3) -> APPENDA_IN_AAA(X1, X2, X3) APPENDA_IN_AAA(X1, .(X1, X2), X3) -> U1_AAA(X1, X2, X3, appendA_in_aaa(X1, X2, X3)) APPENDA_IN_AAA(X1, .(X1, X2), X3) -> APPENDA_IN_AAA(X1, X2, X3) R is empty. The argument filtering Pi contains the following mapping: appendA_in_aaa(x1, x2, x3) = appendA_in_aaa .(x1, x2) = .(x1, x2) PB_IN_AAA(x1, x2, x3) = PB_IN_AAA U2_AAA(x1, x2, x3, x4) = U2_AAA(x4) APPENDA_IN_AAA(x1, x2, x3) = APPENDA_IN_AAA U1_AAA(x1, x2, x3, x4) = U1_AAA(x4) We have to consider all (P,R,Pi)-chains ---------------------------------------- (37) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes. ---------------------------------------- (38) Obligation: Pi DP problem: The TRS P consists of the following rules: APPENDA_IN_AAA(X1, .(X1, X2), X3) -> APPENDA_IN_AAA(X1, X2, X3) R is empty. The argument filtering Pi contains the following mapping: .(x1, x2) = .(x1, x2) APPENDA_IN_AAA(x1, x2, x3) = APPENDA_IN_AAA We have to consider all (P,R,Pi)-chains ---------------------------------------- (39) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (40) Obligation: Q DP problem: The TRS P consists of the following rules: APPENDA_IN_AAA -> APPENDA_IN_AAA R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (41) PrologToIRSwTTransformerProof (SOUND) Transformed Prolog program to IRSwT according to method in Master Thesis of A. Weinert { "root": 3, "program": { "directives": [], "clauses": [ [ "(p X Y Z)", "(append (. X Y) Z Y)" ], [ "(append ([]) Y Y)", null ], [ "(append (. H Xs) Ys (. H Zs))", "(append Xs Ys Zs)" ] ] }, "graph": { "nodes": { "11": { "goal": [{ "clause": 0, "scope": 1, "term": "(p T1 T2 T3)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "3": { "goal": [{ "clause": -1, "scope": -1, "term": "(p T1 T2 T3)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "124": { "goal": [ { "clause": 1, "scope": 2, "term": "(append (. T18 T17) T16 T17)" }, { "clause": 2, "scope": 2, "term": "(append (. T18 T17) T16 T17)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "125": { "goal": [{ "clause": 2, "scope": 2, "term": "(append (. T18 T17) T16 T17)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "126": { "goal": [{ "clause": -1, "scope": -1, "term": "(append (. T33 T32) T31 T32)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "127": { "goal": [], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "119": { "goal": [{ "clause": -1, "scope": -1, "term": "(append (. T18 T17) T16 T17)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "type": "Nodes" }, "edges": [ { "from": 3, "to": 11, "label": "CASE" }, { "from": 11, "to": 119, "label": "ONLY EVAL with clause\np(X9, X10, X11) :- append(.(X9, X10), X11, X10).\nand substitutionT1 -> T18,\nX9 -> T18,\nT2 -> T17,\nX10 -> T17,\nT3 -> T16,\nX11 -> T16,\nT15 -> T16,\nT14 -> T17,\nT13 -> T18" }, { "from": 119, "to": 124, "label": "CASE" }, { "from": 124, "to": 125, "label": "BACKTRACK\nfor clause: append([], Y, Y)because of non-unification" }, { "from": 125, "to": 126, "label": "EVAL with clause\nappend(.(X23, X24), X25, .(X23, X26)) :- append(X24, X25, X26).\nand substitutionT18 -> T33,\nX23 -> T33,\nT17 -> .(T33, T32),\nX24 -> .(T33, T32),\nT16 -> T31,\nX25 -> T31,\nX26 -> T32,\nT28 -> .(T33, T32),\nT29 -> T31,\nT30 -> T32,\nT27 -> T33" }, { "from": 125, "to": 127, "label": "EVAL-BACKTRACK" }, { "from": 126, "to": 119, "label": "INSTANCE with matching:\nT18 -> T33\nT17 -> T32\nT16 -> T31" } ], "type": "Graph" } } ---------------------------------------- (42) Obligation: Rules: f124_in -> f125_in :|: TRUE f125_out -> f124_out :|: TRUE f124_out -> f119_out :|: TRUE f119_in -> f124_in :|: TRUE f126_in -> f119_in :|: TRUE f119_out -> f126_out :|: TRUE f126_out -> f125_out :|: TRUE f127_out -> f125_out :|: TRUE f125_in -> f126_in :|: TRUE f125_in -> f127_in :|: TRUE f3_in -> f11_in :|: TRUE f11_out -> f3_out :|: TRUE f11_in -> f119_in :|: TRUE f119_out -> f11_out :|: TRUE Start term: f3_in ---------------------------------------- (43) IRSwTSimpleDependencyGraphProof (EQUIVALENT) Constructed simple dependency graph. Simplified to the following IRSwTs: intTRSProblem: f124_in -> f125_in :|: TRUE f119_in -> f124_in :|: TRUE f126_in -> f119_in :|: TRUE f125_in -> f126_in :|: TRUE ---------------------------------------- (44) Obligation: Rules: f124_in -> f125_in :|: TRUE f119_in -> f124_in :|: TRUE f126_in -> f119_in :|: TRUE f125_in -> f126_in :|: TRUE ---------------------------------------- (45) IntTRSCompressionProof (EQUIVALENT) Compressed rules. ---------------------------------------- (46) Obligation: Rules: f126_in -> f126_in :|: TRUE ---------------------------------------- (47) IRSFormatTransformerProof (EQUIVALENT) Reformatted IRS to match normalized format (transformed away non-linear left-hand sides, !=, / and %). ---------------------------------------- (48) Obligation: Rules: f126_in -> f126_in :|: TRUE ---------------------------------------- (49) IRSwTTerminationDigraphProof (EQUIVALENT) Constructed termination digraph! Nodes: (1) f126_in -> f126_in :|: TRUE Arcs: (1) -> (1) This digraph is fully evaluated! ---------------------------------------- (50) Obligation: Termination digraph: Nodes: (1) f126_in -> f126_in :|: TRUE Arcs: (1) -> (1) This digraph is fully evaluated! ---------------------------------------- (51) FilterProof (EQUIVALENT) Used the following sort dictionary for filtering: f126_in() Replaced non-predefined constructor symbols by 0. ---------------------------------------- (52) Obligation: Rules: f126_in -> f126_in :|: TRUE ---------------------------------------- (53) IntTRSNonPeriodicNontermProof (COMPLETE) Normalized system to the following form: f(pc) -> f(1) :|: pc = 1 && TRUE Proved unsatisfiability of the following formula, indicating that the system is never left after entering: ((run2_0 = ((1 * 1)) and (((run1_0 * 1)) = ((1 * 1)) and T)) and !(((run2_0 * 1)) = ((1 * 1)) and T)) Proved satisfiability of the following formula, indicating that the system is entered at least once: (run2_0 = ((1 * 1)) and (((run1_0 * 1)) = ((1 * 1)) and T)) ---------------------------------------- (54) NO