3.69/1.73 YES 3.69/1.74 proof of /export/starexec/sandbox/benchmark/theBenchmark.pl 3.69/1.74 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.69/1.74 3.69/1.74 3.69/1.74 Left Termination of the query pattern 3.69/1.74 3.69/1.74 reverse(g,a) 3.69/1.74 3.69/1.74 w.r.t. the given Prolog program could successfully be proven: 3.69/1.74 3.69/1.74 (0) Prolog 3.69/1.74 (1) PrologToPiTRSProof [SOUND, 0 ms] 3.69/1.74 (2) PiTRS 3.69/1.74 (3) DependencyPairsProof [EQUIVALENT, 0 ms] 3.69/1.74 (4) PiDP 3.69/1.74 (5) DependencyGraphProof [EQUIVALENT, 1 ms] 3.69/1.74 (6) PiDP 3.69/1.74 (7) UsableRulesProof [EQUIVALENT, 0 ms] 3.69/1.74 (8) PiDP 3.69/1.74 (9) PiDPToQDPProof [SOUND, 0 ms] 3.69/1.74 (10) QDP 3.69/1.74 (11) QDPSizeChangeProof [EQUIVALENT, 2 ms] 3.69/1.74 (12) YES 3.69/1.74 3.69/1.74 3.69/1.74 ---------------------------------------- 3.69/1.74 3.69/1.74 (0) 3.69/1.74 Obligation: 3.69/1.74 Clauses: 3.69/1.74 3.69/1.74 reverse(X1s, X2s) :- reverse(X1s, [], X2s). 3.69/1.74 reverse([], Xs, Xs). 3.69/1.74 reverse(.(X, X1s), X2s, Ys) :- reverse(X1s, .(X, X2s), Ys). 3.69/1.74 3.69/1.74 3.69/1.74 Query: reverse(g,a) 3.69/1.74 ---------------------------------------- 3.69/1.74 3.69/1.74 (1) PrologToPiTRSProof (SOUND) 3.69/1.74 We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: 3.69/1.74 3.69/1.74 reverse_in_2: (b,f) 3.69/1.74 3.69/1.74 reverse_in_3: (b,b,f) 3.69/1.74 3.69/1.74 Transforming Prolog into the following Term Rewriting System: 3.69/1.74 3.69/1.74 Pi-finite rewrite system: 3.69/1.74 The TRS R consists of the following rules: 3.69/1.74 3.69/1.74 reverse_in_ga(X1s, X2s) -> U1_ga(X1s, X2s, reverse_in_gga(X1s, [], X2s)) 3.69/1.74 reverse_in_gga([], Xs, Xs) -> reverse_out_gga([], Xs, Xs) 3.69/1.74 reverse_in_gga(.(X, X1s), X2s, Ys) -> U2_gga(X, X1s, X2s, Ys, reverse_in_gga(X1s, .(X, X2s), Ys)) 3.69/1.74 U2_gga(X, X1s, X2s, Ys, reverse_out_gga(X1s, .(X, X2s), Ys)) -> reverse_out_gga(.(X, X1s), X2s, Ys) 3.69/1.74 U1_ga(X1s, X2s, reverse_out_gga(X1s, [], X2s)) -> reverse_out_ga(X1s, X2s) 3.69/1.74 3.69/1.74 The argument filtering Pi contains the following mapping: 3.69/1.74 reverse_in_ga(x1, x2) = reverse_in_ga(x1) 3.69/1.74 3.69/1.74 U1_ga(x1, x2, x3) = U1_ga(x3) 3.69/1.74 3.69/1.74 reverse_in_gga(x1, x2, x3) = reverse_in_gga(x1, x2) 3.69/1.74 3.69/1.74 [] = [] 3.69/1.74 3.69/1.74 reverse_out_gga(x1, x2, x3) = reverse_out_gga(x3) 3.69/1.74 3.69/1.74 .(x1, x2) = .(x1, x2) 3.69/1.74 3.69/1.74 U2_gga(x1, x2, x3, x4, x5) = U2_gga(x5) 3.69/1.74 3.69/1.74 reverse_out_ga(x1, x2) = reverse_out_ga(x2) 3.69/1.74 3.69/1.74 3.69/1.74 3.69/1.74 3.69/1.74 3.69/1.74 Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog 3.69/1.74 3.69/1.74 3.69/1.74 3.69/1.74 ---------------------------------------- 3.69/1.74 3.69/1.74 (2) 3.69/1.74 Obligation: 3.69/1.74 Pi-finite rewrite system: 3.69/1.74 The TRS R consists of the following rules: 3.69/1.74 3.69/1.74 reverse_in_ga(X1s, X2s) -> U1_ga(X1s, X2s, reverse_in_gga(X1s, [], X2s)) 3.69/1.74 reverse_in_gga([], Xs, Xs) -> reverse_out_gga([], Xs, Xs) 3.69/1.74 reverse_in_gga(.(X, X1s), X2s, Ys) -> U2_gga(X, X1s, X2s, Ys, reverse_in_gga(X1s, .(X, X2s), Ys)) 3.69/1.74 U2_gga(X, X1s, X2s, Ys, reverse_out_gga(X1s, .(X, X2s), Ys)) -> reverse_out_gga(.(X, X1s), X2s, Ys) 3.69/1.74 U1_ga(X1s, X2s, reverse_out_gga(X1s, [], X2s)) -> reverse_out_ga(X1s, X2s) 3.69/1.74 3.69/1.74 The argument filtering Pi contains the following mapping: 3.69/1.74 reverse_in_ga(x1, x2) = reverse_in_ga(x1) 3.69/1.74 3.69/1.74 U1_ga(x1, x2, x3) = U1_ga(x3) 3.69/1.74 3.69/1.74 reverse_in_gga(x1, x2, x3) = reverse_in_gga(x1, x2) 3.69/1.74 3.69/1.74 [] = [] 3.69/1.74 3.69/1.74 reverse_out_gga(x1, x2, x3) = reverse_out_gga(x3) 3.69/1.74 3.69/1.74 .(x1, x2) = .(x1, x2) 3.69/1.74 3.69/1.74 U2_gga(x1, x2, x3, x4, x5) = U2_gga(x5) 3.69/1.74 3.69/1.74 reverse_out_ga(x1, x2) = reverse_out_ga(x2) 3.69/1.74 3.69/1.74 3.69/1.74 3.69/1.74 ---------------------------------------- 3.69/1.74 3.69/1.74 (3) DependencyPairsProof (EQUIVALENT) 3.69/1.74 Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem: 3.69/1.74 Pi DP problem: 3.69/1.74 The TRS P consists of the following rules: 3.69/1.74 3.69/1.74 REVERSE_IN_GA(X1s, X2s) -> U1_GA(X1s, X2s, reverse_in_gga(X1s, [], X2s)) 3.69/1.74 REVERSE_IN_GA(X1s, X2s) -> REVERSE_IN_GGA(X1s, [], X2s) 3.69/1.74 REVERSE_IN_GGA(.(X, X1s), X2s, Ys) -> U2_GGA(X, X1s, X2s, Ys, reverse_in_gga(X1s, .(X, X2s), Ys)) 3.69/1.74 REVERSE_IN_GGA(.(X, X1s), X2s, Ys) -> REVERSE_IN_GGA(X1s, .(X, X2s), Ys) 3.69/1.74 3.69/1.74 The TRS R consists of the following rules: 3.69/1.74 3.69/1.74 reverse_in_ga(X1s, X2s) -> U1_ga(X1s, X2s, reverse_in_gga(X1s, [], X2s)) 3.69/1.74 reverse_in_gga([], Xs, Xs) -> reverse_out_gga([], Xs, Xs) 3.69/1.74 reverse_in_gga(.(X, X1s), X2s, Ys) -> U2_gga(X, X1s, X2s, Ys, reverse_in_gga(X1s, .(X, X2s), Ys)) 3.69/1.74 U2_gga(X, X1s, X2s, Ys, reverse_out_gga(X1s, .(X, X2s), Ys)) -> reverse_out_gga(.(X, X1s), X2s, Ys) 3.69/1.74 U1_ga(X1s, X2s, reverse_out_gga(X1s, [], X2s)) -> reverse_out_ga(X1s, X2s) 3.69/1.74 3.69/1.74 The argument filtering Pi contains the following mapping: 3.69/1.74 reverse_in_ga(x1, x2) = reverse_in_ga(x1) 3.69/1.74 3.69/1.74 U1_ga(x1, x2, x3) = U1_ga(x3) 3.69/1.74 3.69/1.74 reverse_in_gga(x1, x2, x3) = reverse_in_gga(x1, x2) 3.69/1.74 3.69/1.74 [] = [] 3.69/1.74 3.69/1.74 reverse_out_gga(x1, x2, x3) = reverse_out_gga(x3) 3.69/1.74 3.69/1.74 .(x1, x2) = .(x1, x2) 3.69/1.74 3.69/1.74 U2_gga(x1, x2, x3, x4, x5) = U2_gga(x5) 3.69/1.74 3.69/1.74 reverse_out_ga(x1, x2) = reverse_out_ga(x2) 3.69/1.74 3.69/1.74 REVERSE_IN_GA(x1, x2) = REVERSE_IN_GA(x1) 3.69/1.74 3.69/1.74 U1_GA(x1, x2, x3) = U1_GA(x3) 3.69/1.74 3.69/1.74 REVERSE_IN_GGA(x1, x2, x3) = REVERSE_IN_GGA(x1, x2) 3.69/1.74 3.69/1.74 U2_GGA(x1, x2, x3, x4, x5) = U2_GGA(x5) 3.69/1.74 3.69/1.74 3.69/1.74 We have to consider all (P,R,Pi)-chains 3.69/1.74 ---------------------------------------- 3.69/1.74 3.69/1.74 (4) 3.69/1.74 Obligation: 3.69/1.74 Pi DP problem: 3.69/1.74 The TRS P consists of the following rules: 3.69/1.74 3.69/1.74 REVERSE_IN_GA(X1s, X2s) -> U1_GA(X1s, X2s, reverse_in_gga(X1s, [], X2s)) 3.69/1.74 REVERSE_IN_GA(X1s, X2s) -> REVERSE_IN_GGA(X1s, [], X2s) 3.69/1.74 REVERSE_IN_GGA(.(X, X1s), X2s, Ys) -> U2_GGA(X, X1s, X2s, Ys, reverse_in_gga(X1s, .(X, X2s), Ys)) 3.69/1.74 REVERSE_IN_GGA(.(X, X1s), X2s, Ys) -> REVERSE_IN_GGA(X1s, .(X, X2s), Ys) 3.69/1.74 3.69/1.74 The TRS R consists of the following rules: 3.69/1.74 3.69/1.74 reverse_in_ga(X1s, X2s) -> U1_ga(X1s, X2s, reverse_in_gga(X1s, [], X2s)) 3.69/1.74 reverse_in_gga([], Xs, Xs) -> reverse_out_gga([], Xs, Xs) 3.69/1.74 reverse_in_gga(.(X, X1s), X2s, Ys) -> U2_gga(X, X1s, X2s, Ys, reverse_in_gga(X1s, .(X, X2s), Ys)) 3.69/1.74 U2_gga(X, X1s, X2s, Ys, reverse_out_gga(X1s, .(X, X2s), Ys)) -> reverse_out_gga(.(X, X1s), X2s, Ys) 3.69/1.74 U1_ga(X1s, X2s, reverse_out_gga(X1s, [], X2s)) -> reverse_out_ga(X1s, X2s) 3.69/1.74 3.69/1.74 The argument filtering Pi contains the following mapping: 3.69/1.74 reverse_in_ga(x1, x2) = reverse_in_ga(x1) 3.69/1.74 3.69/1.74 U1_ga(x1, x2, x3) = U1_ga(x3) 3.69/1.74 3.69/1.74 reverse_in_gga(x1, x2, x3) = reverse_in_gga(x1, x2) 3.69/1.74 3.69/1.74 [] = [] 3.69/1.74 3.69/1.74 reverse_out_gga(x1, x2, x3) = reverse_out_gga(x3) 3.69/1.74 3.69/1.74 .(x1, x2) = .(x1, x2) 3.69/1.74 3.69/1.74 U2_gga(x1, x2, x3, x4, x5) = U2_gga(x5) 3.69/1.74 3.69/1.74 reverse_out_ga(x1, x2) = reverse_out_ga(x2) 3.69/1.74 3.69/1.74 REVERSE_IN_GA(x1, x2) = REVERSE_IN_GA(x1) 3.69/1.74 3.69/1.74 U1_GA(x1, x2, x3) = U1_GA(x3) 3.69/1.74 3.69/1.74 REVERSE_IN_GGA(x1, x2, x3) = REVERSE_IN_GGA(x1, x2) 3.69/1.74 3.69/1.74 U2_GGA(x1, x2, x3, x4, x5) = U2_GGA(x5) 3.69/1.74 3.69/1.74 3.69/1.74 We have to consider all (P,R,Pi)-chains 3.69/1.74 ---------------------------------------- 3.69/1.74 3.69/1.74 (5) DependencyGraphProof (EQUIVALENT) 3.69/1.74 The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes. 3.69/1.74 ---------------------------------------- 3.69/1.74 3.69/1.74 (6) 3.69/1.74 Obligation: 3.69/1.74 Pi DP problem: 3.69/1.74 The TRS P consists of the following rules: 3.69/1.74 3.69/1.74 REVERSE_IN_GGA(.(X, X1s), X2s, Ys) -> REVERSE_IN_GGA(X1s, .(X, X2s), Ys) 3.69/1.74 3.69/1.74 The TRS R consists of the following rules: 3.69/1.74 3.69/1.74 reverse_in_ga(X1s, X2s) -> U1_ga(X1s, X2s, reverse_in_gga(X1s, [], X2s)) 3.69/1.74 reverse_in_gga([], Xs, Xs) -> reverse_out_gga([], Xs, Xs) 3.69/1.74 reverse_in_gga(.(X, X1s), X2s, Ys) -> U2_gga(X, X1s, X2s, Ys, reverse_in_gga(X1s, .(X, X2s), Ys)) 3.69/1.74 U2_gga(X, X1s, X2s, Ys, reverse_out_gga(X1s, .(X, X2s), Ys)) -> reverse_out_gga(.(X, X1s), X2s, Ys) 3.69/1.74 U1_ga(X1s, X2s, reverse_out_gga(X1s, [], X2s)) -> reverse_out_ga(X1s, X2s) 3.69/1.74 3.69/1.74 The argument filtering Pi contains the following mapping: 3.69/1.74 reverse_in_ga(x1, x2) = reverse_in_ga(x1) 3.69/1.74 3.69/1.74 U1_ga(x1, x2, x3) = U1_ga(x3) 3.69/1.74 3.69/1.74 reverse_in_gga(x1, x2, x3) = reverse_in_gga(x1, x2) 3.69/1.74 3.69/1.74 [] = [] 3.69/1.74 3.69/1.74 reverse_out_gga(x1, x2, x3) = reverse_out_gga(x3) 3.69/1.74 3.69/1.74 .(x1, x2) = .(x1, x2) 3.69/1.74 3.69/1.74 U2_gga(x1, x2, x3, x4, x5) = U2_gga(x5) 3.69/1.74 3.69/1.74 reverse_out_ga(x1, x2) = reverse_out_ga(x2) 3.69/1.74 3.69/1.74 REVERSE_IN_GGA(x1, x2, x3) = REVERSE_IN_GGA(x1, x2) 3.69/1.74 3.69/1.74 3.69/1.74 We have to consider all (P,R,Pi)-chains 3.69/1.74 ---------------------------------------- 3.69/1.74 3.69/1.74 (7) UsableRulesProof (EQUIVALENT) 3.69/1.74 For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. 3.69/1.74 ---------------------------------------- 3.69/1.74 3.69/1.74 (8) 3.69/1.74 Obligation: 3.69/1.74 Pi DP problem: 3.69/1.74 The TRS P consists of the following rules: 3.69/1.74 3.69/1.74 REVERSE_IN_GGA(.(X, X1s), X2s, Ys) -> REVERSE_IN_GGA(X1s, .(X, X2s), Ys) 3.69/1.74 3.69/1.74 R is empty. 3.69/1.74 The argument filtering Pi contains the following mapping: 3.69/1.74 .(x1, x2) = .(x1, x2) 3.69/1.74 3.69/1.74 REVERSE_IN_GGA(x1, x2, x3) = REVERSE_IN_GGA(x1, x2) 3.69/1.74 3.69/1.74 3.69/1.74 We have to consider all (P,R,Pi)-chains 3.69/1.74 ---------------------------------------- 3.69/1.74 3.69/1.74 (9) PiDPToQDPProof (SOUND) 3.69/1.74 Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. 3.69/1.74 ---------------------------------------- 3.69/1.74 3.69/1.74 (10) 3.69/1.74 Obligation: 3.69/1.74 Q DP problem: 3.69/1.74 The TRS P consists of the following rules: 3.69/1.74 3.69/1.74 REVERSE_IN_GGA(.(X, X1s), X2s) -> REVERSE_IN_GGA(X1s, .(X, X2s)) 3.69/1.74 3.69/1.74 R is empty. 3.69/1.74 Q is empty. 3.69/1.74 We have to consider all (P,Q,R)-chains. 3.69/1.74 ---------------------------------------- 3.69/1.74 3.69/1.74 (11) QDPSizeChangeProof (EQUIVALENT) 3.69/1.74 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. 3.69/1.74 3.69/1.74 From the DPs we obtained the following set of size-change graphs: 3.69/1.74 *REVERSE_IN_GGA(.(X, X1s), X2s) -> REVERSE_IN_GGA(X1s, .(X, X2s)) 3.69/1.74 The graph contains the following edges 1 > 1 3.69/1.74 3.69/1.74 3.69/1.74 ---------------------------------------- 3.69/1.74 3.69/1.74 (12) 3.69/1.74 YES 3.69/1.77 EOF