/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 130 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 0 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 0 ms] (6) QTRS (7) QTRSRRRProof [EQUIVALENT, 20 ms] (8) QTRS (9) QTRSRRRProof [EQUIVALENT, 17 ms] (10) QTRS (11) QTRSRRRProof [EQUIVALENT, 0 ms] (12) QTRS (13) QTRSRRRProof [EQUIVALENT, 10 ms] (14) QTRS (15) QTRSRRRProof [EQUIVALENT, 0 ms] (16) QTRS (17) QTRSRRRProof [EQUIVALENT, 1 ms] (18) QTRS (19) DependencyPairsProof [EQUIVALENT, 3 ms] (20) QDP (21) DependencyGraphProof [EQUIVALENT, 0 ms] (22) AND (23) QDP (24) UsableRulesProof [EQUIVALENT, 0 ms] (25) QDP (26) QDPSizeChangeProof [EQUIVALENT, 0 ms] (27) YES (28) QDP (29) UsableRulesProof [EQUIVALENT, 0 ms] (30) QDP (31) NonTerminationLoopProof [COMPLETE, 0 ms] (32) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: pairNs -> cons(0, n__incr(oddNs)) oddNs -> incr(pairNs) incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS))) take(0, XS) -> nil take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS))) zip(nil, XS) -> nil zip(X, nil) -> nil zip(cons(X, XS), cons(Y, YS)) -> cons(pair(X, Y), n__zip(activate(XS), activate(YS))) tail(cons(X, XS)) -> activate(XS) repItems(nil) -> nil repItems(cons(X, XS)) -> cons(X, n__cons(X, n__repItems(activate(XS)))) incr(X) -> n__incr(X) take(X1, X2) -> n__take(X1, X2) zip(X1, X2) -> n__zip(X1, X2) cons(X1, X2) -> n__cons(X1, X2) repItems(X) -> n__repItems(X) activate(n__incr(X)) -> incr(X) activate(n__take(X1, X2)) -> take(X1, X2) activate(n__zip(X1, X2)) -> zip(X1, X2) activate(n__cons(X1, X2)) -> cons(X1, X2) activate(n__repItems(X)) -> repItems(X) activate(X) -> X Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(activate(x_1)) = 2*x_1 POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2 POL(incr(x_1)) = 2*x_1 POL(n__cons(x_1, x_2)) = x_1 + x_2 POL(n__incr(x_1)) = x_1 POL(n__repItems(x_1)) = x_1 POL(n__take(x_1, x_2)) = x_1 + x_2 POL(n__zip(x_1, x_2)) = 1 + x_1 + x_2 POL(nil) = 0 POL(oddNs) = 0 POL(pair(x_1, x_2)) = x_1 + x_2 POL(pairNs) = 0 POL(repItems(x_1)) = 2*x_1 POL(s(x_1)) = x_1 POL(tail(x_1)) = 2*x_1 POL(take(x_1, x_2)) = 2*x_1 + 2*x_2 POL(zip(x_1, x_2)) = 2 + 2*x_1 + 2*x_2 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: zip(nil, XS) -> nil zip(X, nil) -> nil zip(X1, X2) -> n__zip(X1, X2) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: pairNs -> cons(0, n__incr(oddNs)) oddNs -> incr(pairNs) incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS))) take(0, XS) -> nil take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS))) zip(cons(X, XS), cons(Y, YS)) -> cons(pair(X, Y), n__zip(activate(XS), activate(YS))) tail(cons(X, XS)) -> activate(XS) repItems(nil) -> nil repItems(cons(X, XS)) -> cons(X, n__cons(X, n__repItems(activate(XS)))) incr(X) -> n__incr(X) take(X1, X2) -> n__take(X1, X2) cons(X1, X2) -> n__cons(X1, X2) repItems(X) -> n__repItems(X) activate(n__incr(X)) -> incr(X) activate(n__take(X1, X2)) -> take(X1, X2) activate(n__zip(X1, X2)) -> zip(X1, X2) activate(n__cons(X1, X2)) -> cons(X1, X2) activate(n__repItems(X)) -> repItems(X) activate(X) -> X Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(activate(x_1)) = 2*x_1 POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2 POL(incr(x_1)) = 2*x_1 POL(n__cons(x_1, x_2)) = x_1 + x_2 POL(n__incr(x_1)) = x_1 POL(n__repItems(x_1)) = x_1 POL(n__take(x_1, x_2)) = 2*x_1 + x_2 POL(n__zip(x_1, x_2)) = x_1 + x_2 POL(nil) = 0 POL(oddNs) = 0 POL(pair(x_1, x_2)) = x_1 + 2*x_2 POL(pairNs) = 0 POL(repItems(x_1)) = 2*x_1 POL(s(x_1)) = 2*x_1 POL(tail(x_1)) = 1 + 2*x_1 POL(take(x_1, x_2)) = 2*x_1 + 2*x_2 POL(zip(x_1, x_2)) = 2*x_1 + 2*x_2 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: tail(cons(X, XS)) -> activate(XS) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: pairNs -> cons(0, n__incr(oddNs)) oddNs -> incr(pairNs) incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS))) take(0, XS) -> nil take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS))) zip(cons(X, XS), cons(Y, YS)) -> cons(pair(X, Y), n__zip(activate(XS), activate(YS))) repItems(nil) -> nil repItems(cons(X, XS)) -> cons(X, n__cons(X, n__repItems(activate(XS)))) incr(X) -> n__incr(X) take(X1, X2) -> n__take(X1, X2) cons(X1, X2) -> n__cons(X1, X2) repItems(X) -> n__repItems(X) activate(n__incr(X)) -> incr(X) activate(n__take(X1, X2)) -> take(X1, X2) activate(n__zip(X1, X2)) -> zip(X1, X2) activate(n__cons(X1, X2)) -> cons(X1, X2) activate(n__repItems(X)) -> repItems(X) activate(X) -> X Q is empty. ---------------------------------------- (5) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(activate(x_1)) = 2*x_1 POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2 POL(incr(x_1)) = 2*x_1 POL(n__cons(x_1, x_2)) = x_1 + x_2 POL(n__incr(x_1)) = x_1 POL(n__repItems(x_1)) = 1 + x_1 POL(n__take(x_1, x_2)) = x_1 + x_2 POL(n__zip(x_1, x_2)) = x_1 + x_2 POL(nil) = 0 POL(oddNs) = 0 POL(pair(x_1, x_2)) = 2*x_1 + 2*x_2 POL(pairNs) = 0 POL(repItems(x_1)) = 2 + 2*x_1 POL(s(x_1)) = 2*x_1 POL(take(x_1, x_2)) = 2*x_1 + 2*x_2 POL(zip(x_1, x_2)) = 2*x_1 + 2*x_2 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: repItems(nil) -> nil repItems(X) -> n__repItems(X) ---------------------------------------- (6) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: pairNs -> cons(0, n__incr(oddNs)) oddNs -> incr(pairNs) incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS))) take(0, XS) -> nil take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS))) zip(cons(X, XS), cons(Y, YS)) -> cons(pair(X, Y), n__zip(activate(XS), activate(YS))) repItems(cons(X, XS)) -> cons(X, n__cons(X, n__repItems(activate(XS)))) incr(X) -> n__incr(X) take(X1, X2) -> n__take(X1, X2) cons(X1, X2) -> n__cons(X1, X2) activate(n__incr(X)) -> incr(X) activate(n__take(X1, X2)) -> take(X1, X2) activate(n__zip(X1, X2)) -> zip(X1, X2) activate(n__cons(X1, X2)) -> cons(X1, X2) activate(n__repItems(X)) -> repItems(X) activate(X) -> X Q is empty. ---------------------------------------- (7) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(activate(x_1)) = 2*x_1 POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2 POL(incr(x_1)) = 2*x_1 POL(n__cons(x_1, x_2)) = x_1 + x_2 POL(n__incr(x_1)) = x_1 POL(n__repItems(x_1)) = x_1 POL(n__take(x_1, x_2)) = 1 + x_1 + x_2 POL(n__zip(x_1, x_2)) = x_1 + x_2 POL(nil) = 1 POL(oddNs) = 0 POL(pair(x_1, x_2)) = 2*x_1 + x_2 POL(pairNs) = 0 POL(repItems(x_1)) = 2*x_1 POL(s(x_1)) = 2*x_1 POL(take(x_1, x_2)) = 2 + 2*x_1 + 2*x_2 POL(zip(x_1, x_2)) = 2*x_1 + 2*x_2 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: take(0, XS) -> nil take(X1, X2) -> n__take(X1, X2) ---------------------------------------- (8) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: pairNs -> cons(0, n__incr(oddNs)) oddNs -> incr(pairNs) incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS))) take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS))) zip(cons(X, XS), cons(Y, YS)) -> cons(pair(X, Y), n__zip(activate(XS), activate(YS))) repItems(cons(X, XS)) -> cons(X, n__cons(X, n__repItems(activate(XS)))) incr(X) -> n__incr(X) cons(X1, X2) -> n__cons(X1, X2) activate(n__incr(X)) -> incr(X) activate(n__take(X1, X2)) -> take(X1, X2) activate(n__zip(X1, X2)) -> zip(X1, X2) activate(n__cons(X1, X2)) -> cons(X1, X2) activate(n__repItems(X)) -> repItems(X) activate(X) -> X Q is empty. ---------------------------------------- (9) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(activate(x_1)) = 2*x_1 POL(cons(x_1, x_2)) = 2*x_1 + x_2 POL(incr(x_1)) = 2*x_1 POL(n__cons(x_1, x_2)) = 2*x_1 + x_2 POL(n__incr(x_1)) = x_1 POL(n__repItems(x_1)) = 1 + x_1 POL(n__take(x_1, x_2)) = 1 + x_1 + x_2 POL(n__zip(x_1, x_2)) = x_1 + x_2 POL(oddNs) = 0 POL(pair(x_1, x_2)) = 2*x_1 + x_2 POL(pairNs) = 0 POL(repItems(x_1)) = 2 + 2*x_1 POL(s(x_1)) = 2*x_1 POL(take(x_1, x_2)) = 1 + x_1 + 2*x_2 POL(zip(x_1, x_2)) = 2*x_1 + 2*x_2 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: repItems(cons(X, XS)) -> cons(X, n__cons(X, n__repItems(activate(XS)))) activate(n__take(X1, X2)) -> take(X1, X2) ---------------------------------------- (10) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: pairNs -> cons(0, n__incr(oddNs)) oddNs -> incr(pairNs) incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS))) take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS))) zip(cons(X, XS), cons(Y, YS)) -> cons(pair(X, Y), n__zip(activate(XS), activate(YS))) incr(X) -> n__incr(X) cons(X1, X2) -> n__cons(X1, X2) activate(n__incr(X)) -> incr(X) activate(n__zip(X1, X2)) -> zip(X1, X2) activate(n__cons(X1, X2)) -> cons(X1, X2) activate(n__repItems(X)) -> repItems(X) activate(X) -> X Q is empty. ---------------------------------------- (11) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(activate(x_1)) = x_1 POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2 POL(incr(x_1)) = 2*x_1 POL(n__cons(x_1, x_2)) = 2*x_1 + 2*x_2 POL(n__incr(x_1)) = 2*x_1 POL(n__repItems(x_1)) = 1 + 2*x_1 POL(n__take(x_1, x_2)) = x_1 + 2*x_2 POL(n__zip(x_1, x_2)) = 2*x_1 + 2*x_2 POL(oddNs) = 0 POL(pair(x_1, x_2)) = 2*x_1 + x_2 POL(pairNs) = 0 POL(repItems(x_1)) = 2*x_1 POL(s(x_1)) = 2*x_1 POL(take(x_1, x_2)) = x_1 + 2*x_2 POL(zip(x_1, x_2)) = 2*x_1 + 2*x_2 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: activate(n__repItems(X)) -> repItems(X) ---------------------------------------- (12) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: pairNs -> cons(0, n__incr(oddNs)) oddNs -> incr(pairNs) incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS))) take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS))) zip(cons(X, XS), cons(Y, YS)) -> cons(pair(X, Y), n__zip(activate(XS), activate(YS))) incr(X) -> n__incr(X) cons(X1, X2) -> n__cons(X1, X2) activate(n__incr(X)) -> incr(X) activate(n__zip(X1, X2)) -> zip(X1, X2) activate(n__cons(X1, X2)) -> cons(X1, X2) activate(X) -> X Q is empty. ---------------------------------------- (13) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(activate(x_1)) = x_1 POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2 POL(incr(x_1)) = 2*x_1 POL(n__cons(x_1, x_2)) = 2*x_1 + 2*x_2 POL(n__incr(x_1)) = 2*x_1 POL(n__take(x_1, x_2)) = x_1 + x_2 POL(n__zip(x_1, x_2)) = 2*x_1 + 2*x_2 POL(oddNs) = 0 POL(pair(x_1, x_2)) = 2*x_1 + x_2 POL(pairNs) = 0 POL(s(x_1)) = 2*x_1 POL(take(x_1, x_2)) = 1 + 2*x_1 + x_2 POL(zip(x_1, x_2)) = 2*x_1 + 2*x_2 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS))) ---------------------------------------- (14) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: pairNs -> cons(0, n__incr(oddNs)) oddNs -> incr(pairNs) incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS))) zip(cons(X, XS), cons(Y, YS)) -> cons(pair(X, Y), n__zip(activate(XS), activate(YS))) incr(X) -> n__incr(X) cons(X1, X2) -> n__cons(X1, X2) activate(n__incr(X)) -> incr(X) activate(n__zip(X1, X2)) -> zip(X1, X2) activate(n__cons(X1, X2)) -> cons(X1, X2) activate(X) -> X Q is empty. ---------------------------------------- (15) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(activate(x_1)) = 2*x_1 POL(cons(x_1, x_2)) = 2*x_1 + x_2 POL(incr(x_1)) = 2*x_1 POL(n__cons(x_1, x_2)) = 2*x_1 + x_2 POL(n__incr(x_1)) = x_1 POL(n__zip(x_1, x_2)) = 1 + x_1 + x_2 POL(oddNs) = 0 POL(pair(x_1, x_2)) = 2*x_1 + x_2 POL(pairNs) = 0 POL(s(x_1)) = 2*x_1 POL(zip(x_1, x_2)) = 1 + 2*x_1 + 2*x_2 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: activate(n__zip(X1, X2)) -> zip(X1, X2) ---------------------------------------- (16) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: pairNs -> cons(0, n__incr(oddNs)) oddNs -> incr(pairNs) incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS))) zip(cons(X, XS), cons(Y, YS)) -> cons(pair(X, Y), n__zip(activate(XS), activate(YS))) incr(X) -> n__incr(X) cons(X1, X2) -> n__cons(X1, X2) activate(n__incr(X)) -> incr(X) activate(n__cons(X1, X2)) -> cons(X1, X2) activate(X) -> X Q is empty. ---------------------------------------- (17) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(activate(x_1)) = x_1 POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2 POL(incr(x_1)) = 2*x_1 POL(n__cons(x_1, x_2)) = 2*x_1 + 2*x_2 POL(n__incr(x_1)) = 2*x_1 POL(n__zip(x_1, x_2)) = x_1 + 2*x_2 POL(oddNs) = 0 POL(pair(x_1, x_2)) = x_1 + x_2 POL(pairNs) = 0 POL(s(x_1)) = x_1 POL(zip(x_1, x_2)) = 1 + x_1 + 2*x_2 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: zip(cons(X, XS), cons(Y, YS)) -> cons(pair(X, Y), n__zip(activate(XS), activate(YS))) ---------------------------------------- (18) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: pairNs -> cons(0, n__incr(oddNs)) oddNs -> incr(pairNs) incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS))) incr(X) -> n__incr(X) cons(X1, X2) -> n__cons(X1, X2) activate(n__incr(X)) -> incr(X) activate(n__cons(X1, X2)) -> cons(X1, X2) activate(X) -> X Q is empty. ---------------------------------------- (19) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: PAIRNS -> CONS(0, n__incr(oddNs)) PAIRNS -> ODDNS ODDNS -> INCR(pairNs) ODDNS -> PAIRNS INCR(cons(X, XS)) -> CONS(s(X), n__incr(activate(XS))) INCR(cons(X, XS)) -> ACTIVATE(XS) ACTIVATE(n__incr(X)) -> INCR(X) ACTIVATE(n__cons(X1, X2)) -> CONS(X1, X2) The TRS R consists of the following rules: pairNs -> cons(0, n__incr(oddNs)) oddNs -> incr(pairNs) incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS))) incr(X) -> n__incr(X) cons(X1, X2) -> n__cons(X1, X2) activate(n__incr(X)) -> incr(X) activate(n__cons(X1, X2)) -> cons(X1, X2) activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes. ---------------------------------------- (22) Complex Obligation (AND) ---------------------------------------- (23) Obligation: Q DP problem: The TRS P consists of the following rules: INCR(cons(X, XS)) -> ACTIVATE(XS) ACTIVATE(n__incr(X)) -> INCR(X) The TRS R consists of the following rules: pairNs -> cons(0, n__incr(oddNs)) oddNs -> incr(pairNs) incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS))) incr(X) -> n__incr(X) cons(X1, X2) -> n__cons(X1, X2) activate(n__incr(X)) -> incr(X) activate(n__cons(X1, X2)) -> cons(X1, X2) activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (24) 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. ---------------------------------------- (25) Obligation: Q DP problem: The TRS P consists of the following rules: INCR(cons(X, XS)) -> ACTIVATE(XS) ACTIVATE(n__incr(X)) -> INCR(X) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (26) 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: *ACTIVATE(n__incr(X)) -> INCR(X) The graph contains the following edges 1 > 1 *INCR(cons(X, XS)) -> ACTIVATE(XS) The graph contains the following edges 1 > 1 ---------------------------------------- (27) YES ---------------------------------------- (28) Obligation: Q DP problem: The TRS P consists of the following rules: PAIRNS -> ODDNS ODDNS -> PAIRNS The TRS R consists of the following rules: pairNs -> cons(0, n__incr(oddNs)) oddNs -> incr(pairNs) incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS))) incr(X) -> n__incr(X) cons(X1, X2) -> n__cons(X1, X2) activate(n__incr(X)) -> incr(X) activate(n__cons(X1, X2)) -> cons(X1, X2) activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (29) 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. ---------------------------------------- (30) Obligation: Q DP problem: The TRS P consists of the following rules: PAIRNS -> ODDNS ODDNS -> PAIRNS R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (31) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by narrowing to the left: s = ODDNS evaluates to t =ODDNS Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence ODDNS -> PAIRNS with rule ODDNS -> PAIRNS at position [] and matcher [ ] PAIRNS -> ODDNS with rule PAIRNS -> ODDNS Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence All these steps are and every following step will be a correct step w.r.t to Q. ---------------------------------------- (32) NO