/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern t() w.r.t. the given Prolog program could not be shown: (0) Prolog (1) CutEliminatorProof [SOUND, 0 ms] (2) Prolog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 [EQUIVALENT, 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 [EQUIVALENT, 0 ms] (24) QDP (25) NonTerminationLoopProof [COMPLETE, 0 ms] (26) NO (27) PrologToTRSTransformerProof [SOUND, 0 ms] (28) QTRS (29) QTRSRRRProof [EQUIVALENT, 10 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 [EQUIVALENT, 0 ms] (48) QDP (49) NonTerminationLoopProof [COMPLETE, 0 ms] (50) NO ---------------------------------------- (0) Obligation: Clauses: t :- ','(eq(X, f(X)), !). t :- t. eq(X, X). Query: t() ---------------------------------------- (1) CutEliminatorProof (SOUND) Eliminated all cuts by simply ignoring them[PROLOG]. ---------------------------------------- (2) Obligation: Clauses: t :- eq(X, f(X)). t :- t. eq(X, X). Query: t() ---------------------------------------- (3) PrologToPiTRSProof (SOUND) We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: t_in_ -> U1_(eq_in_aa(X, f(X))) eq_in_aa(X, X) -> eq_out_aa(X, X) U1_(eq_out_aa(X, f(X))) -> t_out_ t_in_ -> U2_(t_in_) U2_(t_out_) -> t_out_ The argument filtering Pi contains the following mapping: t_in_ = t_in_ U1_(x1) = U1_(x1) eq_in_aa(x1, x2) = eq_in_aa eq_out_aa(x1, x2) = eq_out_aa t_out_ = t_out_ U2_(x1) = U2_(x1) Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (4) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: t_in_ -> U1_(eq_in_aa(X, f(X))) eq_in_aa(X, X) -> eq_out_aa(X, X) U1_(eq_out_aa(X, f(X))) -> t_out_ t_in_ -> U2_(t_in_) U2_(t_out_) -> t_out_ The argument filtering Pi contains the following mapping: t_in_ = t_in_ U1_(x1) = U1_(x1) eq_in_aa(x1, x2) = eq_in_aa eq_out_aa(x1, x2) = eq_out_aa t_out_ = t_out_ U2_(x1) = U2_(x1) ---------------------------------------- (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: T_IN_ -> U1_^1(eq_in_aa(X, f(X))) T_IN_ -> EQ_IN_AA(X, f(X)) T_IN_ -> U2_^1(t_in_) T_IN_ -> T_IN_ The TRS R consists of the following rules: t_in_ -> U1_(eq_in_aa(X, f(X))) eq_in_aa(X, X) -> eq_out_aa(X, X) U1_(eq_out_aa(X, f(X))) -> t_out_ t_in_ -> U2_(t_in_) U2_(t_out_) -> t_out_ The argument filtering Pi contains the following mapping: t_in_ = t_in_ U1_(x1) = U1_(x1) eq_in_aa(x1, x2) = eq_in_aa eq_out_aa(x1, x2) = eq_out_aa t_out_ = t_out_ U2_(x1) = U2_(x1) T_IN_ = T_IN_ U1_^1(x1) = U1_^1(x1) EQ_IN_AA(x1, x2) = EQ_IN_AA U2_^1(x1) = U2_^1(x1) We have to consider all (P,R,Pi)-chains ---------------------------------------- (6) Obligation: Pi DP problem: The TRS P consists of the following rules: T_IN_ -> U1_^1(eq_in_aa(X, f(X))) T_IN_ -> EQ_IN_AA(X, f(X)) T_IN_ -> U2_^1(t_in_) T_IN_ -> T_IN_ The TRS R consists of the following rules: t_in_ -> U1_(eq_in_aa(X, f(X))) eq_in_aa(X, X) -> eq_out_aa(X, X) U1_(eq_out_aa(X, f(X))) -> t_out_ t_in_ -> U2_(t_in_) U2_(t_out_) -> t_out_ The argument filtering Pi contains the following mapping: t_in_ = t_in_ U1_(x1) = U1_(x1) eq_in_aa(x1, x2) = eq_in_aa eq_out_aa(x1, x2) = eq_out_aa t_out_ = t_out_ U2_(x1) = U2_(x1) T_IN_ = T_IN_ U1_^1(x1) = U1_^1(x1) EQ_IN_AA(x1, x2) = EQ_IN_AA U2_^1(x1) = U2_^1(x1) 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: T_IN_ -> T_IN_ The TRS R consists of the following rules: t_in_ -> U1_(eq_in_aa(X, f(X))) eq_in_aa(X, X) -> eq_out_aa(X, X) U1_(eq_out_aa(X, f(X))) -> t_out_ t_in_ -> U2_(t_in_) U2_(t_out_) -> t_out_ The argument filtering Pi contains the following mapping: t_in_ = t_in_ U1_(x1) = U1_(x1) eq_in_aa(x1, x2) = eq_in_aa eq_out_aa(x1, x2) = eq_out_aa t_out_ = t_out_ U2_(x1) = U2_(x1) T_IN_ = T_IN_ 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: T_IN_ -> T_IN_ R is empty. Pi is empty. We have to consider all (P,R,Pi)-chains ---------------------------------------- (11) PiDPToQDPProof (EQUIVALENT) 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: T_IN_ -> T_IN_ 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 = T_IN_ evaluates to t =T_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 T_IN_ to T_IN_. ---------------------------------------- (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: Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: t_in_ -> U1_(eq_in_aa(X, f(X))) eq_in_aa(X, X) -> eq_out_aa(X, X) U1_(eq_out_aa(X, f(X))) -> t_out_ t_in_ -> U2_(t_in_) U2_(t_out_) -> t_out_ The argument filtering Pi contains the following mapping: t_in_ = t_in_ U1_(x1) = U1_(x1) eq_in_aa(x1, x2) = eq_in_aa eq_out_aa(x1, x2) = eq_out_aa t_out_ = t_out_ U2_(x1) = U2_(x1) Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (16) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: t_in_ -> U1_(eq_in_aa(X, f(X))) eq_in_aa(X, X) -> eq_out_aa(X, X) U1_(eq_out_aa(X, f(X))) -> t_out_ t_in_ -> U2_(t_in_) U2_(t_out_) -> t_out_ The argument filtering Pi contains the following mapping: t_in_ = t_in_ U1_(x1) = U1_(x1) eq_in_aa(x1, x2) = eq_in_aa eq_out_aa(x1, x2) = eq_out_aa t_out_ = t_out_ U2_(x1) = U2_(x1) ---------------------------------------- (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: T_IN_ -> U1_^1(eq_in_aa(X, f(X))) T_IN_ -> EQ_IN_AA(X, f(X)) T_IN_ -> U2_^1(t_in_) T_IN_ -> T_IN_ The TRS R consists of the following rules: t_in_ -> U1_(eq_in_aa(X, f(X))) eq_in_aa(X, X) -> eq_out_aa(X, X) U1_(eq_out_aa(X, f(X))) -> t_out_ t_in_ -> U2_(t_in_) U2_(t_out_) -> t_out_ The argument filtering Pi contains the following mapping: t_in_ = t_in_ U1_(x1) = U1_(x1) eq_in_aa(x1, x2) = eq_in_aa eq_out_aa(x1, x2) = eq_out_aa t_out_ = t_out_ U2_(x1) = U2_(x1) T_IN_ = T_IN_ U1_^1(x1) = U1_^1(x1) EQ_IN_AA(x1, x2) = EQ_IN_AA U2_^1(x1) = U2_^1(x1) We have to consider all (P,R,Pi)-chains ---------------------------------------- (18) Obligation: Pi DP problem: The TRS P consists of the following rules: T_IN_ -> U1_^1(eq_in_aa(X, f(X))) T_IN_ -> EQ_IN_AA(X, f(X)) T_IN_ -> U2_^1(t_in_) T_IN_ -> T_IN_ The TRS R consists of the following rules: t_in_ -> U1_(eq_in_aa(X, f(X))) eq_in_aa(X, X) -> eq_out_aa(X, X) U1_(eq_out_aa(X, f(X))) -> t_out_ t_in_ -> U2_(t_in_) U2_(t_out_) -> t_out_ The argument filtering Pi contains the following mapping: t_in_ = t_in_ U1_(x1) = U1_(x1) eq_in_aa(x1, x2) = eq_in_aa eq_out_aa(x1, x2) = eq_out_aa t_out_ = t_out_ U2_(x1) = U2_(x1) T_IN_ = T_IN_ U1_^1(x1) = U1_^1(x1) EQ_IN_AA(x1, x2) = EQ_IN_AA U2_^1(x1) = U2_^1(x1) 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: T_IN_ -> T_IN_ The TRS R consists of the following rules: t_in_ -> U1_(eq_in_aa(X, f(X))) eq_in_aa(X, X) -> eq_out_aa(X, X) U1_(eq_out_aa(X, f(X))) -> t_out_ t_in_ -> U2_(t_in_) U2_(t_out_) -> t_out_ The argument filtering Pi contains the following mapping: t_in_ = t_in_ U1_(x1) = U1_(x1) eq_in_aa(x1, x2) = eq_in_aa eq_out_aa(x1, x2) = eq_out_aa t_out_ = t_out_ U2_(x1) = U2_(x1) T_IN_ = T_IN_ 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: T_IN_ -> T_IN_ R is empty. Pi is empty. We have to consider all (P,R,Pi)-chains ---------------------------------------- (23) PiDPToQDPProof (EQUIVALENT) 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: T_IN_ -> T_IN_ 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 = T_IN_ evaluates to t =T_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 T_IN_ to T_IN_. ---------------------------------------- (26) NO ---------------------------------------- (27) PrologToTRSTransformerProof (SOUND) Transformed Prolog program to TRS. { "root": 9, "program": { "directives": [], "clauses": [ [ "(t)", "(',' (eq X (f X)) (!))" ], [ "(t)", "(t)" ], [ "(eq X X)", null ] ] }, "graph": { "nodes": { "44": { "goal": [ { "clause": -1, "scope": -1, "term": "(',' (eq X2 (f X2)) (!_1))" }, { "clause": 1, "scope": 1, "term": "(t)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": ["X2"], "exprvars": [] } }, "45": { "goal": [ { "clause": 2, "scope": 2, "term": "(',' (eq X2 (f X2)) (!_1))" }, { "clause": -1, "scope": 2, "term": null }, { "clause": 1, "scope": 1, "term": "(t)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": ["X2"], "exprvars": [] } }, "46": { "goal": [ { "clause": -1, "scope": 2, "term": null }, { "clause": 1, "scope": 1, "term": "(t)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "47": { "goal": [{ "clause": 1, "scope": 1, "term": "(t)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "48": { "goal": [{ "clause": -1, "scope": -1, "term": "(t)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "9": { "goal": [{ "clause": -1, "scope": -1, "term": "(t)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "type": "Nodes", "10": { "goal": [ { "clause": 0, "scope": 1, "term": "(t)" }, { "clause": 1, "scope": 1, "term": "(t)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } } }, "edges": [ { "from": 9, "to": 10, "label": "CASE" }, { "from": 10, "to": 44, "label": "ONLY EVAL with clause\nt :- ','(eq(X2, f(X2)), !_1).\nand substitution" }, { "from": 44, "to": 45, "label": "CASE" }, { "from": 45, "to": 46, "label": "BACKTRACK\nfor clause: eq(X, X)because of non-unification" }, { "from": 46, "to": 47, "label": "FAILURE" }, { "from": 47, "to": 48, "label": "ONLY EVAL with clause\nt :- t.\nand substitution" }, { "from": 48, "to": 9, "label": "INSTANCE" } ], "type": "Graph" } } ---------------------------------------- (28) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f9_in -> U1(f9_in) U1(f9_out1) -> f9_out1 Q is empty. ---------------------------------------- (29) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(U1(x_1)) = 2*x_1 POL(f9_in) = 0 POL(f9_out1) = 1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: U1(f9_out1) -> f9_out1 ---------------------------------------- (30) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f9_in -> U1(f9_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: f9_in -> U1(f9_in) The set Q consists of the following terms: f9_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: F9_IN -> F9_IN The TRS R consists of the following rules: f9_in -> U1(f9_in) The set Q consists of the following terms: f9_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: F9_IN -> F9_IN R is empty. The set Q consists of the following terms: f9_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]. f9_in ---------------------------------------- (38) Obligation: Q DP problem: The TRS P consists of the following rules: F9_IN -> F9_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 = F9_IN evaluates to t =F9_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 F9_IN to F9_IN. ---------------------------------------- (40) NO ---------------------------------------- (41) PrologToDTProblemTransformerProof (SOUND) Built DT problem from termination graph DT10. { "root": 1, "program": { "directives": [], "clauses": [ [ "(t)", "(',' (eq X (f X)) (!))" ], [ "(t)", "(t)" ], [ "(eq X X)", null ] ] }, "graph": { "nodes": { "1": { "goal": [{ "clause": -1, "scope": -1, "term": "(t)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "2": { "goal": [ { "clause": 0, "scope": 1, "term": "(t)" }, { "clause": 1, "scope": 1, "term": "(t)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "39": { "goal": [ { "clause": -1, "scope": -1, "term": "(',' (eq X1 (f X1)) (!_1))" }, { "clause": 1, "scope": 1, "term": "(t)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": ["X1"], "exprvars": [] } }, "type": "Nodes", "40": { "goal": [ { "clause": 2, "scope": 2, "term": "(',' (eq X1 (f X1)) (!_1))" }, { "clause": -1, "scope": 2, "term": null }, { "clause": 1, "scope": 1, "term": "(t)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": ["X1"], "exprvars": [] } }, "41": { "goal": [ { "clause": -1, "scope": 2, "term": null }, { "clause": 1, "scope": 1, "term": "(t)" } ], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "42": { "goal": [{ "clause": 1, "scope": 1, "term": "(t)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } }, "43": { "goal": [{ "clause": -1, "scope": -1, "term": "(t)" }], "kb": { "nonunifying": [], "intvars": {}, "arithmetic": { "type": "PlainIntegerRelationState", "relations": [] }, "ground": [], "free": [], "exprvars": [] } } }, "edges": [ { "from": 1, "to": 2, "label": "CASE" }, { "from": 2, "to": 39, "label": "ONLY EVAL with clause\nt :- ','(eq(X1, f(X1)), !_1).\nand substitution" }, { "from": 39, "to": 40, "label": "CASE" }, { "from": 40, "to": 41, "label": "BACKTRACK\nfor clause: eq(X, X)because of non-unification" }, { "from": 41, "to": 42, "label": "FAILURE" }, { "from": 42, "to": 43, "label": "ONLY EVAL with clause\nt :- t.\nand substitution" }, { "from": 43, "to": 1, "label": "INSTANCE" } ], "type": "Graph" } } ---------------------------------------- (42) Obligation: Triples: tA :- tA. Clauses: tcA :- tcA. Afs: tA = tA ---------------------------------------- (43) TriplesToPiDPProof (SOUND) We use the technique of [DT09]. With regard to the inferred argument filtering the predicates were used in the following modes: Transforming TRIPLES into the following Term Rewriting System: Pi DP problem: The TRS P consists of the following rules: TA_IN_ -> U1_^1(tA_in_) TA_IN_ -> TA_IN_ 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 ---------------------------------------- (44) Obligation: Pi DP problem: The TRS P consists of the following rules: TA_IN_ -> U1_^1(tA_in_) TA_IN_ -> TA_IN_ R is empty. Pi is empty. 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: TA_IN_ -> TA_IN_ R is empty. Pi is empty. We have to consider all (P,R,Pi)-chains ---------------------------------------- (47) PiDPToQDPProof (EQUIVALENT) 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: TA_IN_ -> TA_IN_ 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 = TA_IN_ evaluates to t =TA_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 TA_IN_ to TA_IN_. ---------------------------------------- (50) NO