4.02/1.81 YES 4.06/1.81 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 4.06/1.81 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.06/1.81 4.06/1.81 4.06/1.81 Termination w.r.t. Q of the given QTRS could be proven: 4.06/1.81 4.06/1.81 (0) QTRS 4.06/1.81 (1) QTRSRRRProof [EQUIVALENT, 65 ms] 4.06/1.81 (2) QTRS 4.06/1.81 (3) QTRSRRRProof [EQUIVALENT, 14 ms] 4.06/1.81 (4) QTRS 4.06/1.81 (5) DependencyPairsProof [EQUIVALENT, 16 ms] 4.06/1.81 (6) QDP 4.06/1.81 (7) DependencyGraphProof [EQUIVALENT, 0 ms] 4.06/1.81 (8) AND 4.06/1.81 (9) QDP 4.06/1.81 (10) UsableRulesProof [EQUIVALENT, 0 ms] 4.06/1.81 (11) QDP 4.06/1.81 (12) QReductionProof [EQUIVALENT, 0 ms] 4.06/1.81 (13) QDP 4.06/1.81 (14) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.06/1.81 (15) YES 4.06/1.81 (16) QDP 4.06/1.81 (17) UsableRulesProof [EQUIVALENT, 0 ms] 4.06/1.81 (18) QDP 4.06/1.81 (19) QReductionProof [EQUIVALENT, 0 ms] 4.06/1.81 (20) QDP 4.06/1.81 (21) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.06/1.81 (22) YES 4.06/1.81 4.06/1.81 4.06/1.81 ---------------------------------------- 4.06/1.81 4.06/1.81 (0) 4.06/1.81 Obligation: 4.06/1.81 Q restricted rewrite system: 4.06/1.81 The TRS R consists of the following rules: 4.06/1.81 4.06/1.81 intlist(nil) -> nil 4.06/1.81 intlist(cons(x, y)) -> cons(s(x), intlist(y)) 4.06/1.81 int(0, 0) -> cons(0, nil) 4.06/1.81 int(0, s(y)) -> cons(0, int(s(0), s(y))) 4.06/1.81 int(s(x), 0) -> nil 4.06/1.81 int(s(x), s(y)) -> intlist(int(x, y)) 4.06/1.81 4.06/1.81 The set Q consists of the following terms: 4.06/1.81 4.06/1.81 intlist(nil) 4.06/1.81 intlist(cons(x0, x1)) 4.06/1.81 int(0, 0) 4.06/1.81 int(0, s(x0)) 4.06/1.81 int(s(x0), 0) 4.06/1.81 int(s(x0), s(x1)) 4.06/1.81 4.06/1.81 4.06/1.81 ---------------------------------------- 4.06/1.81 4.06/1.81 (1) QTRSRRRProof (EQUIVALENT) 4.06/1.81 Used ordering: 4.06/1.81 Polynomial interpretation [POLO]: 4.06/1.81 4.06/1.81 POL(0) = 0 4.06/1.81 POL(cons(x_1, x_2)) = 2*x_1 + x_2 4.06/1.81 POL(int(x_1, x_2)) = 1 + 2*x_1 + 2*x_2 4.06/1.81 POL(intlist(x_1)) = x_1 4.06/1.81 POL(nil) = 0 4.06/1.81 POL(s(x_1)) = x_1 4.06/1.81 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 4.06/1.81 4.06/1.81 int(0, 0) -> cons(0, nil) 4.06/1.81 int(s(x), 0) -> nil 4.06/1.81 4.06/1.81 4.06/1.81 4.06/1.81 4.06/1.81 ---------------------------------------- 4.06/1.81 4.06/1.81 (2) 4.06/1.81 Obligation: 4.06/1.81 Q restricted rewrite system: 4.06/1.81 The TRS R consists of the following rules: 4.06/1.81 4.06/1.81 intlist(nil) -> nil 4.06/1.81 intlist(cons(x, y)) -> cons(s(x), intlist(y)) 4.06/1.81 int(0, s(y)) -> cons(0, int(s(0), s(y))) 4.06/1.81 int(s(x), s(y)) -> intlist(int(x, y)) 4.06/1.81 4.06/1.81 The set Q consists of the following terms: 4.06/1.81 4.06/1.81 intlist(nil) 4.06/1.81 intlist(cons(x0, x1)) 4.06/1.81 int(0, 0) 4.06/1.81 int(0, s(x0)) 4.06/1.81 int(s(x0), 0) 4.06/1.81 int(s(x0), s(x1)) 4.06/1.81 4.06/1.81 4.06/1.81 ---------------------------------------- 4.06/1.81 4.06/1.81 (3) QTRSRRRProof (EQUIVALENT) 4.06/1.81 Used ordering: 4.06/1.81 Polynomial interpretation [POLO]: 4.06/1.81 4.06/1.81 POL(0) = 0 4.06/1.81 POL(cons(x_1, x_2)) = 2*x_1 + x_2 4.06/1.81 POL(int(x_1, x_2)) = 2*x_1 + x_2 4.06/1.81 POL(intlist(x_1)) = 2*x_1 4.06/1.81 POL(nil) = 1 4.06/1.81 POL(s(x_1)) = 2*x_1 4.06/1.81 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 4.06/1.81 4.06/1.81 intlist(nil) -> nil 4.06/1.81 4.06/1.81 4.06/1.81 4.06/1.81 4.06/1.81 ---------------------------------------- 4.06/1.81 4.06/1.81 (4) 4.06/1.81 Obligation: 4.06/1.81 Q restricted rewrite system: 4.06/1.81 The TRS R consists of the following rules: 4.06/1.81 4.06/1.81 intlist(cons(x, y)) -> cons(s(x), intlist(y)) 4.06/1.81 int(0, s(y)) -> cons(0, int(s(0), s(y))) 4.06/1.81 int(s(x), s(y)) -> intlist(int(x, y)) 4.06/1.81 4.06/1.81 The set Q consists of the following terms: 4.06/1.81 4.06/1.81 intlist(nil) 4.06/1.81 intlist(cons(x0, x1)) 4.06/1.81 int(0, 0) 4.06/1.81 int(0, s(x0)) 4.06/1.81 int(s(x0), 0) 4.06/1.81 int(s(x0), s(x1)) 4.06/1.81 4.06/1.81 4.06/1.81 ---------------------------------------- 4.06/1.81 4.06/1.81 (5) DependencyPairsProof (EQUIVALENT) 4.06/1.81 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 4.06/1.81 ---------------------------------------- 4.06/1.81 4.06/1.81 (6) 4.06/1.81 Obligation: 4.06/1.81 Q DP problem: 4.06/1.81 The TRS P consists of the following rules: 4.06/1.81 4.06/1.81 INTLIST(cons(x, y)) -> INTLIST(y) 4.06/1.81 INT(0, s(y)) -> INT(s(0), s(y)) 4.06/1.81 INT(s(x), s(y)) -> INTLIST(int(x, y)) 4.06/1.81 INT(s(x), s(y)) -> INT(x, y) 4.06/1.81 4.06/1.81 The TRS R consists of the following rules: 4.06/1.81 4.06/1.81 intlist(cons(x, y)) -> cons(s(x), intlist(y)) 4.06/1.81 int(0, s(y)) -> cons(0, int(s(0), s(y))) 4.06/1.81 int(s(x), s(y)) -> intlist(int(x, y)) 4.06/1.81 4.06/1.81 The set Q consists of the following terms: 4.06/1.81 4.06/1.81 intlist(nil) 4.06/1.81 intlist(cons(x0, x1)) 4.06/1.81 int(0, 0) 4.06/1.81 int(0, s(x0)) 4.06/1.81 int(s(x0), 0) 4.06/1.81 int(s(x0), s(x1)) 4.06/1.81 4.06/1.81 We have to consider all minimal (P,Q,R)-chains. 4.06/1.81 ---------------------------------------- 4.06/1.81 4.06/1.81 (7) DependencyGraphProof (EQUIVALENT) 4.06/1.81 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node. 4.06/1.81 ---------------------------------------- 4.06/1.82 4.06/1.82 (8) 4.06/1.82 Complex Obligation (AND) 4.06/1.82 4.06/1.82 ---------------------------------------- 4.06/1.82 4.06/1.82 (9) 4.06/1.82 Obligation: 4.06/1.82 Q DP problem: 4.06/1.82 The TRS P consists of the following rules: 4.06/1.82 4.06/1.82 INTLIST(cons(x, y)) -> INTLIST(y) 4.06/1.82 4.06/1.82 The TRS R consists of the following rules: 4.06/1.82 4.06/1.82 intlist(cons(x, y)) -> cons(s(x), intlist(y)) 4.06/1.82 int(0, s(y)) -> cons(0, int(s(0), s(y))) 4.06/1.82 int(s(x), s(y)) -> intlist(int(x, y)) 4.06/1.82 4.06/1.82 The set Q consists of the following terms: 4.06/1.82 4.06/1.82 intlist(nil) 4.06/1.82 intlist(cons(x0, x1)) 4.06/1.82 int(0, 0) 4.06/1.82 int(0, s(x0)) 4.06/1.82 int(s(x0), 0) 4.06/1.82 int(s(x0), s(x1)) 4.06/1.82 4.06/1.82 We have to consider all minimal (P,Q,R)-chains. 4.06/1.82 ---------------------------------------- 4.06/1.82 4.06/1.82 (10) UsableRulesProof (EQUIVALENT) 4.06/1.82 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. 4.06/1.82 ---------------------------------------- 4.06/1.82 4.06/1.82 (11) 4.06/1.82 Obligation: 4.06/1.82 Q DP problem: 4.06/1.82 The TRS P consists of the following rules: 4.06/1.82 4.06/1.82 INTLIST(cons(x, y)) -> INTLIST(y) 4.06/1.82 4.06/1.82 R is empty. 4.06/1.82 The set Q consists of the following terms: 4.06/1.82 4.06/1.82 intlist(nil) 4.06/1.82 intlist(cons(x0, x1)) 4.06/1.82 int(0, 0) 4.06/1.82 int(0, s(x0)) 4.06/1.82 int(s(x0), 0) 4.06/1.82 int(s(x0), s(x1)) 4.06/1.82 4.06/1.82 We have to consider all minimal (P,Q,R)-chains. 4.06/1.82 ---------------------------------------- 4.06/1.82 4.06/1.82 (12) QReductionProof (EQUIVALENT) 4.06/1.82 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 4.06/1.82 4.06/1.82 intlist(nil) 4.06/1.82 intlist(cons(x0, x1)) 4.06/1.82 int(0, 0) 4.06/1.82 int(0, s(x0)) 4.06/1.82 int(s(x0), 0) 4.06/1.82 int(s(x0), s(x1)) 4.06/1.82 4.06/1.82 4.06/1.82 ---------------------------------------- 4.06/1.82 4.06/1.82 (13) 4.06/1.82 Obligation: 4.06/1.82 Q DP problem: 4.06/1.82 The TRS P consists of the following rules: 4.06/1.82 4.06/1.82 INTLIST(cons(x, y)) -> INTLIST(y) 4.06/1.82 4.06/1.82 R is empty. 4.06/1.82 Q is empty. 4.06/1.82 We have to consider all minimal (P,Q,R)-chains. 4.06/1.82 ---------------------------------------- 4.06/1.82 4.06/1.82 (14) QDPSizeChangeProof (EQUIVALENT) 4.06/1.82 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. 4.06/1.82 4.06/1.82 From the DPs we obtained the following set of size-change graphs: 4.06/1.82 *INTLIST(cons(x, y)) -> INTLIST(y) 4.06/1.82 The graph contains the following edges 1 > 1 4.06/1.82 4.06/1.82 4.06/1.82 ---------------------------------------- 4.06/1.82 4.06/1.82 (15) 4.06/1.82 YES 4.06/1.82 4.06/1.82 ---------------------------------------- 4.06/1.82 4.06/1.82 (16) 4.06/1.82 Obligation: 4.06/1.82 Q DP problem: 4.06/1.82 The TRS P consists of the following rules: 4.06/1.82 4.06/1.82 INT(s(x), s(y)) -> INT(x, y) 4.06/1.82 INT(0, s(y)) -> INT(s(0), s(y)) 4.06/1.82 4.06/1.82 The TRS R consists of the following rules: 4.06/1.82 4.06/1.82 intlist(cons(x, y)) -> cons(s(x), intlist(y)) 4.06/1.82 int(0, s(y)) -> cons(0, int(s(0), s(y))) 4.06/1.82 int(s(x), s(y)) -> intlist(int(x, y)) 4.06/1.82 4.06/1.82 The set Q consists of the following terms: 4.06/1.82 4.06/1.82 intlist(nil) 4.06/1.82 intlist(cons(x0, x1)) 4.06/1.82 int(0, 0) 4.06/1.82 int(0, s(x0)) 4.06/1.82 int(s(x0), 0) 4.06/1.82 int(s(x0), s(x1)) 4.06/1.82 4.06/1.82 We have to consider all minimal (P,Q,R)-chains. 4.06/1.82 ---------------------------------------- 4.06/1.82 4.06/1.82 (17) UsableRulesProof (EQUIVALENT) 4.06/1.82 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. 4.06/1.82 ---------------------------------------- 4.06/1.82 4.06/1.82 (18) 4.06/1.82 Obligation: 4.06/1.82 Q DP problem: 4.06/1.82 The TRS P consists of the following rules: 4.06/1.82 4.06/1.82 INT(s(x), s(y)) -> INT(x, y) 4.06/1.82 INT(0, s(y)) -> INT(s(0), s(y)) 4.06/1.82 4.06/1.82 R is empty. 4.06/1.82 The set Q consists of the following terms: 4.06/1.82 4.06/1.82 intlist(nil) 4.06/1.82 intlist(cons(x0, x1)) 4.06/1.82 int(0, 0) 4.06/1.82 int(0, s(x0)) 4.06/1.82 int(s(x0), 0) 4.06/1.82 int(s(x0), s(x1)) 4.06/1.82 4.06/1.82 We have to consider all minimal (P,Q,R)-chains. 4.06/1.82 ---------------------------------------- 4.06/1.82 4.06/1.82 (19) QReductionProof (EQUIVALENT) 4.06/1.82 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 4.06/1.82 4.06/1.82 intlist(nil) 4.06/1.82 intlist(cons(x0, x1)) 4.06/1.82 int(0, 0) 4.06/1.82 int(0, s(x0)) 4.06/1.82 int(s(x0), 0) 4.06/1.82 int(s(x0), s(x1)) 4.06/1.82 4.06/1.82 4.06/1.82 ---------------------------------------- 4.06/1.82 4.06/1.82 (20) 4.06/1.82 Obligation: 4.06/1.82 Q DP problem: 4.06/1.82 The TRS P consists of the following rules: 4.06/1.82 4.06/1.82 INT(s(x), s(y)) -> INT(x, y) 4.06/1.82 INT(0, s(y)) -> INT(s(0), s(y)) 4.06/1.82 4.06/1.82 R is empty. 4.06/1.82 Q is empty. 4.06/1.82 We have to consider all minimal (P,Q,R)-chains. 4.06/1.82 ---------------------------------------- 4.06/1.82 4.06/1.82 (21) QDPSizeChangeProof (EQUIVALENT) 4.06/1.82 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. 4.06/1.82 4.06/1.82 From the DPs we obtained the following set of size-change graphs: 4.06/1.82 *INT(s(x), s(y)) -> INT(x, y) 4.06/1.82 The graph contains the following edges 1 > 1, 2 > 2 4.06/1.82 4.06/1.82 4.06/1.82 *INT(0, s(y)) -> INT(s(0), s(y)) 4.06/1.82 The graph contains the following edges 2 >= 2 4.06/1.82 4.06/1.82 4.06/1.82 ---------------------------------------- 4.06/1.82 4.06/1.82 (22) 4.06/1.82 YES 4.07/1.85 EOF