YES proof of Mixed_TRS_Ex1_Luc04b_GM.trs # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 53 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 0 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 7 ms] (6) QTRS (7) DependencyPairsProof [EQUIVALENT, 0 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) QDP (11) MRRProof [EQUIVALENT, 0 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) QDP (15) QDPOrderProof [EQUIVALENT, 24 ms] (16) QDP (17) UsableRulesProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPSizeChangeProof [EQUIVALENT, 0 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__nats -> cons(0, incr(nats)) a__pairs -> cons(0, incr(odds)) a__odds -> a__incr(a__pairs) a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS)) a__head(cons(X, XS)) -> mark(X) a__tail(cons(X, XS)) -> mark(XS) mark(nats) -> a__nats mark(pairs) -> a__pairs mark(odds) -> a__odds mark(incr(X)) -> a__incr(mark(X)) mark(head(X)) -> a__head(mark(X)) mark(tail(X)) -> a__tail(mark(X)) mark(0) -> 0 mark(s(X)) -> s(mark(X)) mark(nil) -> nil mark(cons(X1, X2)) -> cons(mark(X1), X2) a__nats -> nats a__pairs -> pairs a__odds -> odds a__incr(X) -> incr(X) a__head(X) -> head(X) a__tail(X) -> tail(X) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a__head(x_1)) = 2 + x_1 POL(a__incr(x_1)) = 2*x_1 POL(a__nats) = 2 POL(a__odds) = 0 POL(a__pairs) = 0 POL(a__tail(x_1)) = 1 + 2*x_1 POL(cons(x_1, x_2)) = 2*x_1 + x_2 POL(head(x_1)) = 1 + x_1 POL(incr(x_1)) = 2*x_1 POL(mark(x_1)) = 2*x_1 POL(nats) = 1 POL(nil) = 1 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: a__head(cons(X, XS)) -> mark(X) a__tail(cons(X, XS)) -> mark(XS) mark(tail(X)) -> a__tail(mark(X)) mark(nil) -> nil a__nats -> nats a__head(X) -> head(X) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__nats -> cons(0, incr(nats)) a__pairs -> cons(0, incr(odds)) a__odds -> a__incr(a__pairs) a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS)) mark(nats) -> a__nats mark(pairs) -> a__pairs mark(odds) -> a__odds mark(incr(X)) -> a__incr(mark(X)) mark(head(X)) -> a__head(mark(X)) mark(0) -> 0 mark(s(X)) -> s(mark(X)) mark(cons(X1, X2)) -> cons(mark(X1), X2) a__pairs -> pairs a__odds -> odds a__incr(X) -> incr(X) a__tail(X) -> tail(X) Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a__head(x_1)) = x_1 POL(a__incr(x_1)) = 2*x_1 POL(a__nats) = 0 POL(a__odds) = 0 POL(a__pairs) = 0 POL(a__tail(x_1)) = 2 + 2*x_1 POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2 POL(head(x_1)) = 2*x_1 POL(incr(x_1)) = 2*x_1 POL(mark(x_1)) = 2*x_1 POL(nats) = 0 POL(odds) = 0 POL(pairs) = 0 POL(s(x_1)) = x_1 POL(tail(x_1)) = 1 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: a__tail(X) -> tail(X) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__nats -> cons(0, incr(nats)) a__pairs -> cons(0, incr(odds)) a__odds -> a__incr(a__pairs) a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS)) mark(nats) -> a__nats mark(pairs) -> a__pairs mark(odds) -> a__odds mark(incr(X)) -> a__incr(mark(X)) mark(head(X)) -> a__head(mark(X)) mark(0) -> 0 mark(s(X)) -> s(mark(X)) mark(cons(X1, X2)) -> cons(mark(X1), X2) a__pairs -> pairs a__odds -> odds a__incr(X) -> incr(X) Q is empty. ---------------------------------------- (5) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a__head(x_1)) = x_1 POL(a__incr(x_1)) = 2*x_1 POL(a__nats) = 0 POL(a__odds) = 0 POL(a__pairs) = 0 POL(cons(x_1, x_2)) = x_1 + 2*x_2 POL(head(x_1)) = 1 + x_1 POL(incr(x_1)) = 2*x_1 POL(mark(x_1)) = x_1 POL(nats) = 0 POL(odds) = 0 POL(pairs) = 0 POL(s(x_1)) = x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: mark(head(X)) -> a__head(mark(X)) ---------------------------------------- (6) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__nats -> cons(0, incr(nats)) a__pairs -> cons(0, incr(odds)) a__odds -> a__incr(a__pairs) a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS)) mark(nats) -> a__nats mark(pairs) -> a__pairs mark(odds) -> a__odds mark(incr(X)) -> a__incr(mark(X)) mark(0) -> 0 mark(s(X)) -> s(mark(X)) mark(cons(X1, X2)) -> cons(mark(X1), X2) a__pairs -> pairs a__odds -> odds a__incr(X) -> incr(X) Q is empty. ---------------------------------------- (7) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: A__ODDS -> A__INCR(a__pairs) A__ODDS -> A__PAIRS A__INCR(cons(X, XS)) -> MARK(X) MARK(nats) -> A__NATS MARK(pairs) -> A__PAIRS MARK(odds) -> A__ODDS MARK(incr(X)) -> A__INCR(mark(X)) MARK(incr(X)) -> MARK(X) MARK(s(X)) -> MARK(X) MARK(cons(X1, X2)) -> MARK(X1) The TRS R consists of the following rules: a__nats -> cons(0, incr(nats)) a__pairs -> cons(0, incr(odds)) a__odds -> a__incr(a__pairs) a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS)) mark(nats) -> a__nats mark(pairs) -> a__pairs mark(odds) -> a__odds mark(incr(X)) -> a__incr(mark(X)) mark(0) -> 0 mark(s(X)) -> s(mark(X)) mark(cons(X1, X2)) -> cons(mark(X1), X2) a__pairs -> pairs a__odds -> odds a__incr(X) -> incr(X) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: A__INCR(cons(X, XS)) -> MARK(X) MARK(odds) -> A__ODDS A__ODDS -> A__INCR(a__pairs) MARK(incr(X)) -> A__INCR(mark(X)) MARK(incr(X)) -> MARK(X) MARK(s(X)) -> MARK(X) MARK(cons(X1, X2)) -> MARK(X1) The TRS R consists of the following rules: a__nats -> cons(0, incr(nats)) a__pairs -> cons(0, incr(odds)) a__odds -> a__incr(a__pairs) a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS)) mark(nats) -> a__nats mark(pairs) -> a__pairs mark(odds) -> a__odds mark(incr(X)) -> a__incr(mark(X)) mark(0) -> 0 mark(s(X)) -> s(mark(X)) mark(cons(X1, X2)) -> cons(mark(X1), X2) a__pairs -> pairs a__odds -> odds a__incr(X) -> incr(X) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: MARK(odds) -> A__ODDS Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(A__INCR(x_1)) = x_1 POL(A__ODDS) = 2 POL(MARK(x_1)) = 2*x_1 POL(a__incr(x_1)) = x_1 POL(a__nats) = 2 POL(a__odds) = 2 POL(a__pairs) = 2 POL(cons(x_1, x_2)) = 2*x_1 + x_2 POL(incr(x_1)) = x_1 POL(mark(x_1)) = x_1 POL(nats) = 2 POL(odds) = 2 POL(pairs) = 2 POL(s(x_1)) = x_1 ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: A__INCR(cons(X, XS)) -> MARK(X) A__ODDS -> A__INCR(a__pairs) MARK(incr(X)) -> A__INCR(mark(X)) MARK(incr(X)) -> MARK(X) MARK(s(X)) -> MARK(X) MARK(cons(X1, X2)) -> MARK(X1) The TRS R consists of the following rules: a__nats -> cons(0, incr(nats)) a__pairs -> cons(0, incr(odds)) a__odds -> a__incr(a__pairs) a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS)) mark(nats) -> a__nats mark(pairs) -> a__pairs mark(odds) -> a__odds mark(incr(X)) -> a__incr(mark(X)) mark(0) -> 0 mark(s(X)) -> s(mark(X)) mark(cons(X1, X2)) -> cons(mark(X1), X2) a__pairs -> pairs a__odds -> odds a__incr(X) -> incr(X) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(incr(X)) -> A__INCR(mark(X)) A__INCR(cons(X, XS)) -> MARK(X) MARK(incr(X)) -> MARK(X) MARK(s(X)) -> MARK(X) MARK(cons(X1, X2)) -> MARK(X1) The TRS R consists of the following rules: a__nats -> cons(0, incr(nats)) a__pairs -> cons(0, incr(odds)) a__odds -> a__incr(a__pairs) a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS)) mark(nats) -> a__nats mark(pairs) -> a__pairs mark(odds) -> a__odds mark(incr(X)) -> a__incr(mark(X)) mark(0) -> 0 mark(s(X)) -> s(mark(X)) mark(cons(X1, X2)) -> cons(mark(X1), X2) a__pairs -> pairs a__odds -> odds a__incr(X) -> incr(X) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. MARK(incr(X)) -> A__INCR(mark(X)) A__INCR(cons(X, XS)) -> MARK(X) MARK(cons(X1, X2)) -> MARK(X1) The remaining pairs can at least be oriented weakly. Used ordering: Combined order from the following AFS and order. MARK(x1) = MARK(x1) incr(x1) = x1 A__INCR(x1) = A__INCR(x1) mark(x1) = x1 cons(x1, x2) = cons(x1) s(x1) = x1 nats = nats a__nats = a__nats pairs = pairs a__pairs = a__pairs odds = odds a__odds = a__odds a__incr(x1) = x1 0 = 0 Recursive path order with status [RPO]. Quasi-Precedence: nats > a__nats > [MARK_1, cons_1] > A__INCR_1 nats > a__nats > 0 [pairs, a__pairs, odds, a__odds] > [MARK_1, cons_1] > A__INCR_1 [pairs, a__pairs, odds, a__odds] > 0 Status: MARK_1: [1] A__INCR_1: multiset status cons_1: [1] nats: multiset status a__nats: multiset status pairs: multiset status a__pairs: multiset status odds: multiset status a__odds: multiset status 0: multiset status The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: mark(nats) -> a__nats mark(pairs) -> a__pairs mark(odds) -> a__odds mark(incr(X)) -> a__incr(mark(X)) mark(0) -> 0 mark(s(X)) -> s(mark(X)) mark(cons(X1, X2)) -> cons(mark(X1), X2) a__odds -> odds a__odds -> a__incr(a__pairs) a__pairs -> cons(0, incr(odds)) a__pairs -> pairs a__incr(X) -> incr(X) a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS)) a__nats -> cons(0, incr(nats)) ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(incr(X)) -> MARK(X) MARK(s(X)) -> MARK(X) The TRS R consists of the following rules: a__nats -> cons(0, incr(nats)) a__pairs -> cons(0, incr(odds)) a__odds -> a__incr(a__pairs) a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS)) mark(nats) -> a__nats mark(pairs) -> a__pairs mark(odds) -> a__odds mark(incr(X)) -> a__incr(mark(X)) mark(0) -> 0 mark(s(X)) -> s(mark(X)) mark(cons(X1, X2)) -> cons(mark(X1), X2) a__pairs -> pairs a__odds -> odds a__incr(X) -> incr(X) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) 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. ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(incr(X)) -> MARK(X) MARK(s(X)) -> MARK(X) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) 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: *MARK(incr(X)) -> MARK(X) The graph contains the following edges 1 > 1 *MARK(s(X)) -> MARK(X) The graph contains the following edges 1 > 1 ---------------------------------------- (20) YES