3.44/1.68 YES 3.44/1.69 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 3.44/1.69 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.44/1.69 3.44/1.69 3.44/1.69 Termination w.r.t. Q of the given QTRS could be proven: 3.44/1.69 3.44/1.69 (0) QTRS 3.44/1.69 (1) DependencyPairsProof [EQUIVALENT, 0 ms] 3.44/1.69 (2) QDP 3.44/1.69 (3) DependencyGraphProof [EQUIVALENT, 0 ms] 3.44/1.69 (4) QDP 3.44/1.69 (5) UsableRulesProof [EQUIVALENT, 0 ms] 3.44/1.69 (6) QDP 3.44/1.69 (7) QReductionProof [EQUIVALENT, 0 ms] 3.44/1.69 (8) QDP 3.44/1.69 (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] 3.44/1.69 (10) YES 3.44/1.69 3.44/1.69 3.44/1.69 ---------------------------------------- 3.44/1.69 3.44/1.69 (0) 3.44/1.69 Obligation: 3.44/1.69 Q restricted rewrite system: 3.44/1.69 The TRS R consists of the following rules: 3.44/1.69 3.44/1.69 a__f(X, g(X), Y) -> a__f(Y, Y, Y) 3.44/1.69 a__g(b) -> c 3.44/1.69 a__b -> c 3.44/1.69 mark(f(X1, X2, X3)) -> a__f(X1, X2, X3) 3.44/1.69 mark(g(X)) -> a__g(mark(X)) 3.44/1.69 mark(b) -> a__b 3.44/1.69 mark(c) -> c 3.44/1.69 a__f(X1, X2, X3) -> f(X1, X2, X3) 3.44/1.69 a__g(X) -> g(X) 3.44/1.69 a__b -> b 3.44/1.69 3.44/1.69 The set Q consists of the following terms: 3.44/1.69 3.44/1.69 a__b 3.44/1.69 mark(f(x0, x1, x2)) 3.44/1.69 mark(g(x0)) 3.44/1.69 mark(b) 3.44/1.69 mark(c) 3.44/1.69 a__f(x0, x1, x2) 3.44/1.69 a__g(x0) 3.44/1.69 3.44/1.69 3.44/1.69 ---------------------------------------- 3.44/1.69 3.44/1.69 (1) DependencyPairsProof (EQUIVALENT) 3.44/1.69 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 3.44/1.69 ---------------------------------------- 3.44/1.69 3.44/1.69 (2) 3.44/1.69 Obligation: 3.44/1.69 Q DP problem: 3.44/1.69 The TRS P consists of the following rules: 3.44/1.69 3.44/1.69 A__F(X, g(X), Y) -> A__F(Y, Y, Y) 3.44/1.69 MARK(f(X1, X2, X3)) -> A__F(X1, X2, X3) 3.44/1.69 MARK(g(X)) -> A__G(mark(X)) 3.44/1.69 MARK(g(X)) -> MARK(X) 3.44/1.69 MARK(b) -> A__B 3.44/1.69 3.44/1.69 The TRS R consists of the following rules: 3.44/1.69 3.44/1.69 a__f(X, g(X), Y) -> a__f(Y, Y, Y) 3.44/1.69 a__g(b) -> c 3.44/1.69 a__b -> c 3.44/1.69 mark(f(X1, X2, X3)) -> a__f(X1, X2, X3) 3.44/1.69 mark(g(X)) -> a__g(mark(X)) 3.44/1.69 mark(b) -> a__b 3.44/1.69 mark(c) -> c 3.44/1.69 a__f(X1, X2, X3) -> f(X1, X2, X3) 3.44/1.69 a__g(X) -> g(X) 3.44/1.69 a__b -> b 3.44/1.69 3.44/1.69 The set Q consists of the following terms: 3.44/1.69 3.44/1.69 a__b 3.44/1.69 mark(f(x0, x1, x2)) 3.44/1.69 mark(g(x0)) 3.44/1.69 mark(b) 3.44/1.69 mark(c) 3.44/1.69 a__f(x0, x1, x2) 3.44/1.69 a__g(x0) 3.44/1.69 3.44/1.69 We have to consider all minimal (P,Q,R)-chains. 3.44/1.69 ---------------------------------------- 3.44/1.69 3.44/1.69 (3) DependencyGraphProof (EQUIVALENT) 3.44/1.69 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes. 3.44/1.69 ---------------------------------------- 3.44/1.69 3.44/1.69 (4) 3.44/1.69 Obligation: 3.44/1.69 Q DP problem: 3.44/1.69 The TRS P consists of the following rules: 3.44/1.69 3.44/1.69 MARK(g(X)) -> MARK(X) 3.44/1.69 3.44/1.69 The TRS R consists of the following rules: 3.44/1.69 3.44/1.69 a__f(X, g(X), Y) -> a__f(Y, Y, Y) 3.44/1.69 a__g(b) -> c 3.44/1.69 a__b -> c 3.44/1.69 mark(f(X1, X2, X3)) -> a__f(X1, X2, X3) 3.44/1.69 mark(g(X)) -> a__g(mark(X)) 3.44/1.69 mark(b) -> a__b 3.44/1.69 mark(c) -> c 3.44/1.69 a__f(X1, X2, X3) -> f(X1, X2, X3) 3.44/1.69 a__g(X) -> g(X) 3.44/1.69 a__b -> b 3.44/1.69 3.44/1.69 The set Q consists of the following terms: 3.44/1.69 3.44/1.69 a__b 3.44/1.69 mark(f(x0, x1, x2)) 3.44/1.69 mark(g(x0)) 3.44/1.69 mark(b) 3.44/1.69 mark(c) 3.44/1.69 a__f(x0, x1, x2) 3.44/1.69 a__g(x0) 3.44/1.69 3.44/1.69 We have to consider all minimal (P,Q,R)-chains. 3.44/1.69 ---------------------------------------- 3.44/1.69 3.44/1.69 (5) UsableRulesProof (EQUIVALENT) 3.44/1.69 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. 3.44/1.69 ---------------------------------------- 3.44/1.69 3.44/1.69 (6) 3.44/1.69 Obligation: 3.44/1.69 Q DP problem: 3.44/1.69 The TRS P consists of the following rules: 3.44/1.69 3.44/1.69 MARK(g(X)) -> MARK(X) 3.44/1.69 3.44/1.69 R is empty. 3.44/1.69 The set Q consists of the following terms: 3.44/1.69 3.44/1.69 a__b 3.44/1.69 mark(f(x0, x1, x2)) 3.44/1.69 mark(g(x0)) 3.44/1.69 mark(b) 3.44/1.69 mark(c) 3.44/1.69 a__f(x0, x1, x2) 3.44/1.69 a__g(x0) 3.44/1.69 3.44/1.69 We have to consider all minimal (P,Q,R)-chains. 3.44/1.69 ---------------------------------------- 3.44/1.69 3.44/1.69 (7) QReductionProof (EQUIVALENT) 3.44/1.69 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 3.44/1.69 3.44/1.69 a__b 3.44/1.69 mark(f(x0, x1, x2)) 3.44/1.69 mark(g(x0)) 3.44/1.69 mark(b) 3.44/1.69 mark(c) 3.44/1.69 a__f(x0, x1, x2) 3.44/1.69 a__g(x0) 3.44/1.69 3.44/1.69 3.44/1.69 ---------------------------------------- 3.44/1.69 3.44/1.69 (8) 3.44/1.69 Obligation: 3.44/1.69 Q DP problem: 3.44/1.69 The TRS P consists of the following rules: 3.44/1.69 3.44/1.69 MARK(g(X)) -> MARK(X) 3.44/1.69 3.44/1.69 R is empty. 3.44/1.69 Q is empty. 3.44/1.69 We have to consider all minimal (P,Q,R)-chains. 3.44/1.69 ---------------------------------------- 3.44/1.69 3.44/1.69 (9) QDPSizeChangeProof (EQUIVALENT) 3.44/1.69 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. 3.44/1.69 3.44/1.69 From the DPs we obtained the following set of size-change graphs: 3.44/1.69 *MARK(g(X)) -> MARK(X) 3.44/1.69 The graph contains the following edges 1 > 1 3.44/1.69 3.44/1.69 3.44/1.69 ---------------------------------------- 3.44/1.69 3.44/1.69 (10) 3.44/1.69 YES 3.57/1.72 EOF