NO proof of Transformed_CSR_04_Ex1_Luc04b_Z.trs # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 61 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) NonTerminationLoopProof [COMPLETE, 0 ms] (16) NO (17) QDP (18) UsableRulesProof [EQUIVALENT, 0 ms] (19) QDP (20) NonTerminationLoopProof [COMPLETE, 0 ms] (21) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: nats -> cons(0, n__incr(nats)) pairs -> cons(0, n__incr(odds)) odds -> incr(pairs) incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS))) head(cons(X, XS)) -> X tail(cons(X, XS)) -> activate(XS) incr(X) -> n__incr(X) activate(n__incr(X)) -> incr(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)) = x_1 + x_2 POL(head(x_1)) = 2 + x_1 POL(incr(x_1)) = 2*x_1 POL(n__incr(x_1)) = x_1 POL(nats) = 0 POL(odds) = 0 POL(pairs) = 0 POL(s(x_1)) = x_1 POL(tail(x_1)) = 1 + 2*x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: head(cons(X, XS)) -> X tail(cons(X, XS)) -> activate(XS) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: nats -> cons(0, n__incr(nats)) pairs -> cons(0, n__incr(odds)) odds -> incr(pairs) incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS))) incr(X) -> n__incr(X) activate(n__incr(X)) -> incr(X) activate(X) -> X Q is empty. ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: NATS -> NATS PAIRS -> ODDS ODDS -> INCR(pairs) ODDS -> PAIRS INCR(cons(X, XS)) -> ACTIVATE(XS) ACTIVATE(n__incr(X)) -> INCR(X) The TRS R consists of the following rules: nats -> cons(0, n__incr(nats)) pairs -> cons(0, n__incr(odds)) odds -> incr(pairs) incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS))) incr(X) -> n__incr(X) activate(n__incr(X)) -> incr(X) activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 1 less node. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVATE(n__incr(X)) -> INCR(X) INCR(cons(X, XS)) -> ACTIVATE(XS) The TRS R consists of the following rules: nats -> cons(0, n__incr(nats)) pairs -> cons(0, n__incr(odds)) odds -> incr(pairs) incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS))) incr(X) -> n__incr(X) activate(n__incr(X)) -> incr(X) activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) 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. ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVATE(n__incr(X)) -> INCR(X) INCR(cons(X, XS)) -> ACTIVATE(XS) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) 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: *INCR(cons(X, XS)) -> ACTIVATE(XS) The graph contains the following edges 1 > 1 *ACTIVATE(n__incr(X)) -> INCR(X) The graph contains the following edges 1 > 1 ---------------------------------------- (11) YES ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: ODDS -> PAIRS PAIRS -> ODDS The TRS R consists of the following rules: nats -> cons(0, n__incr(nats)) pairs -> cons(0, n__incr(odds)) odds -> incr(pairs) incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS))) incr(X) -> n__incr(X) activate(n__incr(X)) -> incr(X) activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) 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. ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: ODDS -> PAIRS PAIRS -> ODDS R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) 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 = PAIRS evaluates to t =PAIRS Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence PAIRS -> ODDS with rule PAIRS -> ODDS at position [] and matcher [ ] ODDS -> PAIRS with rule ODDS -> PAIRS 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. ---------------------------------------- (16) NO ---------------------------------------- (17) Obligation: Q DP problem: The TRS P consists of the following rules: NATS -> NATS The TRS R consists of the following rules: nats -> cons(0, n__incr(nats)) pairs -> cons(0, n__incr(odds)) odds -> incr(pairs) incr(cons(X, XS)) -> cons(s(X), n__incr(activate(XS))) incr(X) -> n__incr(X) activate(n__incr(X)) -> incr(X) activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (18) 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. ---------------------------------------- (19) Obligation: Q DP problem: The TRS P consists of the following rules: NATS -> NATS R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (20) 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 = NATS evaluates to t =NATS 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 NATS to NATS. ---------------------------------------- (21) NO