3.67/1.80 YES 3.67/1.81 proof of /export/starexec/sandbox/benchmark/theBenchmark.pl 3.67/1.81 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.67/1.81 3.67/1.81 3.67/1.81 Left Termination of the query pattern 3.67/1.81 3.67/1.81 avg(g,a,g) 3.67/1.81 3.67/1.81 w.r.t. the given Prolog program could successfully be proven: 3.67/1.81 3.67/1.81 (0) Prolog 3.67/1.81 (1) PrologToPiTRSProof [SOUND, 0 ms] 3.67/1.81 (2) PiTRS 3.67/1.81 (3) DependencyPairsProof [EQUIVALENT, 0 ms] 3.67/1.81 (4) PiDP 3.67/1.81 (5) DependencyGraphProof [EQUIVALENT, 0 ms] 3.67/1.81 (6) PiDP 3.67/1.81 (7) UsableRulesProof [EQUIVALENT, 0 ms] 3.67/1.81 (8) PiDP 3.67/1.81 (9) PiDPToQDPProof [SOUND, 0 ms] 3.67/1.81 (10) QDP 3.67/1.81 (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] 3.67/1.81 (12) YES 3.67/1.81 3.67/1.81 3.67/1.81 ---------------------------------------- 3.67/1.81 3.67/1.81 (0) 3.67/1.81 Obligation: 3.67/1.81 Clauses: 3.67/1.81 3.67/1.81 avg(s(X), Y, Z) :- avg(X, s(Y), Z). 3.67/1.81 avg(X, s(s(s(Y))), s(Z)) :- avg(s(X), Y, Z). 3.67/1.81 avg(0, 0, 0). 3.67/1.81 avg(0, s(0), 0). 3.67/1.81 avg(0, s(s(0)), s(0)). 3.67/1.81 3.67/1.81 3.67/1.81 Query: avg(g,a,g) 3.67/1.81 ---------------------------------------- 3.67/1.81 3.67/1.81 (1) PrologToPiTRSProof (SOUND) 3.67/1.81 We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: 3.67/1.81 3.67/1.81 avg_in_3: (b,f,b) 3.67/1.81 3.67/1.81 Transforming Prolog into the following Term Rewriting System: 3.67/1.81 3.67/1.81 Pi-finite rewrite system: 3.67/1.81 The TRS R consists of the following rules: 3.67/1.81 3.67/1.81 avg_in_gag(s(X), Y, Z) -> U1_gag(X, Y, Z, avg_in_gag(X, s(Y), Z)) 3.67/1.81 avg_in_gag(X, s(s(s(Y))), s(Z)) -> U2_gag(X, Y, Z, avg_in_gag(s(X), Y, Z)) 3.67/1.81 avg_in_gag(0, 0, 0) -> avg_out_gag(0, 0, 0) 3.67/1.81 avg_in_gag(0, s(0), 0) -> avg_out_gag(0, s(0), 0) 3.67/1.81 avg_in_gag(0, s(s(0)), s(0)) -> avg_out_gag(0, s(s(0)), s(0)) 3.67/1.81 U2_gag(X, Y, Z, avg_out_gag(s(X), Y, Z)) -> avg_out_gag(X, s(s(s(Y))), s(Z)) 3.67/1.81 U1_gag(X, Y, Z, avg_out_gag(X, s(Y), Z)) -> avg_out_gag(s(X), Y, Z) 3.67/1.81 3.67/1.81 The argument filtering Pi contains the following mapping: 3.67/1.81 avg_in_gag(x1, x2, x3) = avg_in_gag(x1, x3) 3.67/1.81 3.67/1.81 s(x1) = s(x1) 3.67/1.81 3.67/1.81 U1_gag(x1, x2, x3, x4) = U1_gag(x4) 3.67/1.81 3.67/1.81 U2_gag(x1, x2, x3, x4) = U2_gag(x4) 3.67/1.81 3.67/1.81 0 = 0 3.67/1.81 3.67/1.81 avg_out_gag(x1, x2, x3) = avg_out_gag(x2) 3.67/1.81 3.67/1.81 3.67/1.81 3.67/1.81 3.67/1.81 3.67/1.81 Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog 3.67/1.81 3.67/1.81 3.67/1.81 3.67/1.81 ---------------------------------------- 3.67/1.81 3.67/1.81 (2) 3.67/1.81 Obligation: 3.67/1.81 Pi-finite rewrite system: 3.67/1.81 The TRS R consists of the following rules: 3.67/1.81 3.67/1.81 avg_in_gag(s(X), Y, Z) -> U1_gag(X, Y, Z, avg_in_gag(X, s(Y), Z)) 3.67/1.81 avg_in_gag(X, s(s(s(Y))), s(Z)) -> U2_gag(X, Y, Z, avg_in_gag(s(X), Y, Z)) 3.67/1.81 avg_in_gag(0, 0, 0) -> avg_out_gag(0, 0, 0) 3.67/1.81 avg_in_gag(0, s(0), 0) -> avg_out_gag(0, s(0), 0) 3.67/1.81 avg_in_gag(0, s(s(0)), s(0)) -> avg_out_gag(0, s(s(0)), s(0)) 3.67/1.81 U2_gag(X, Y, Z, avg_out_gag(s(X), Y, Z)) -> avg_out_gag(X, s(s(s(Y))), s(Z)) 3.67/1.81 U1_gag(X, Y, Z, avg_out_gag(X, s(Y), Z)) -> avg_out_gag(s(X), Y, Z) 3.67/1.81 3.67/1.81 The argument filtering Pi contains the following mapping: 3.67/1.81 avg_in_gag(x1, x2, x3) = avg_in_gag(x1, x3) 3.67/1.81 3.67/1.81 s(x1) = s(x1) 3.67/1.81 3.67/1.81 U1_gag(x1, x2, x3, x4) = U1_gag(x4) 3.67/1.81 3.67/1.81 U2_gag(x1, x2, x3, x4) = U2_gag(x4) 3.67/1.81 3.67/1.81 0 = 0 3.67/1.81 3.67/1.81 avg_out_gag(x1, x2, x3) = avg_out_gag(x2) 3.67/1.81 3.67/1.81 3.67/1.81 3.67/1.81 ---------------------------------------- 3.67/1.81 3.67/1.81 (3) DependencyPairsProof (EQUIVALENT) 3.67/1.81 Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem: 3.67/1.81 Pi DP problem: 3.67/1.81 The TRS P consists of the following rules: 3.67/1.81 3.67/1.81 AVG_IN_GAG(s(X), Y, Z) -> U1_GAG(X, Y, Z, avg_in_gag(X, s(Y), Z)) 3.67/1.81 AVG_IN_GAG(s(X), Y, Z) -> AVG_IN_GAG(X, s(Y), Z) 3.67/1.81 AVG_IN_GAG(X, s(s(s(Y))), s(Z)) -> U2_GAG(X, Y, Z, avg_in_gag(s(X), Y, Z)) 3.67/1.81 AVG_IN_GAG(X, s(s(s(Y))), s(Z)) -> AVG_IN_GAG(s(X), Y, Z) 3.67/1.81 3.67/1.81 The TRS R consists of the following rules: 3.67/1.81 3.67/1.81 avg_in_gag(s(X), Y, Z) -> U1_gag(X, Y, Z, avg_in_gag(X, s(Y), Z)) 3.67/1.81 avg_in_gag(X, s(s(s(Y))), s(Z)) -> U2_gag(X, Y, Z, avg_in_gag(s(X), Y, Z)) 3.67/1.81 avg_in_gag(0, 0, 0) -> avg_out_gag(0, 0, 0) 3.67/1.81 avg_in_gag(0, s(0), 0) -> avg_out_gag(0, s(0), 0) 3.67/1.81 avg_in_gag(0, s(s(0)), s(0)) -> avg_out_gag(0, s(s(0)), s(0)) 3.67/1.81 U2_gag(X, Y, Z, avg_out_gag(s(X), Y, Z)) -> avg_out_gag(X, s(s(s(Y))), s(Z)) 3.67/1.81 U1_gag(X, Y, Z, avg_out_gag(X, s(Y), Z)) -> avg_out_gag(s(X), Y, Z) 3.67/1.81 3.67/1.81 The argument filtering Pi contains the following mapping: 3.67/1.81 avg_in_gag(x1, x2, x3) = avg_in_gag(x1, x3) 3.67/1.81 3.67/1.81 s(x1) = s(x1) 3.67/1.81 3.67/1.81 U1_gag(x1, x2, x3, x4) = U1_gag(x4) 3.67/1.81 3.67/1.81 U2_gag(x1, x2, x3, x4) = U2_gag(x4) 3.67/1.81 3.67/1.81 0 = 0 3.67/1.81 3.67/1.81 avg_out_gag(x1, x2, x3) = avg_out_gag(x2) 3.67/1.81 3.67/1.81 AVG_IN_GAG(x1, x2, x3) = AVG_IN_GAG(x1, x3) 3.67/1.81 3.67/1.81 U1_GAG(x1, x2, x3, x4) = U1_GAG(x4) 3.67/1.81 3.67/1.81 U2_GAG(x1, x2, x3, x4) = U2_GAG(x4) 3.67/1.81 3.67/1.81 3.67/1.81 We have to consider all (P,R,Pi)-chains 3.67/1.81 ---------------------------------------- 3.67/1.81 3.67/1.81 (4) 3.67/1.81 Obligation: 3.67/1.81 Pi DP problem: 3.67/1.81 The TRS P consists of the following rules: 3.67/1.81 3.67/1.81 AVG_IN_GAG(s(X), Y, Z) -> U1_GAG(X, Y, Z, avg_in_gag(X, s(Y), Z)) 3.67/1.81 AVG_IN_GAG(s(X), Y, Z) -> AVG_IN_GAG(X, s(Y), Z) 3.67/1.81 AVG_IN_GAG(X, s(s(s(Y))), s(Z)) -> U2_GAG(X, Y, Z, avg_in_gag(s(X), Y, Z)) 3.67/1.81 AVG_IN_GAG(X, s(s(s(Y))), s(Z)) -> AVG_IN_GAG(s(X), Y, Z) 3.67/1.81 3.67/1.81 The TRS R consists of the following rules: 3.67/1.81 3.67/1.81 avg_in_gag(s(X), Y, Z) -> U1_gag(X, Y, Z, avg_in_gag(X, s(Y), Z)) 3.67/1.81 avg_in_gag(X, s(s(s(Y))), s(Z)) -> U2_gag(X, Y, Z, avg_in_gag(s(X), Y, Z)) 3.67/1.81 avg_in_gag(0, 0, 0) -> avg_out_gag(0, 0, 0) 3.67/1.81 avg_in_gag(0, s(0), 0) -> avg_out_gag(0, s(0), 0) 3.67/1.81 avg_in_gag(0, s(s(0)), s(0)) -> avg_out_gag(0, s(s(0)), s(0)) 3.67/1.81 U2_gag(X, Y, Z, avg_out_gag(s(X), Y, Z)) -> avg_out_gag(X, s(s(s(Y))), s(Z)) 3.67/1.81 U1_gag(X, Y, Z, avg_out_gag(X, s(Y), Z)) -> avg_out_gag(s(X), Y, Z) 3.67/1.81 3.67/1.81 The argument filtering Pi contains the following mapping: 3.67/1.81 avg_in_gag(x1, x2, x3) = avg_in_gag(x1, x3) 3.67/1.81 3.67/1.81 s(x1) = s(x1) 3.67/1.81 3.67/1.81 U1_gag(x1, x2, x3, x4) = U1_gag(x4) 3.67/1.81 3.67/1.81 U2_gag(x1, x2, x3, x4) = U2_gag(x4) 3.67/1.81 3.67/1.81 0 = 0 3.67/1.81 3.67/1.81 avg_out_gag(x1, x2, x3) = avg_out_gag(x2) 3.67/1.81 3.67/1.81 AVG_IN_GAG(x1, x2, x3) = AVG_IN_GAG(x1, x3) 3.67/1.81 3.67/1.81 U1_GAG(x1, x2, x3, x4) = U1_GAG(x4) 3.67/1.81 3.67/1.81 U2_GAG(x1, x2, x3, x4) = U2_GAG(x4) 3.67/1.81 3.67/1.81 3.67/1.81 We have to consider all (P,R,Pi)-chains 3.67/1.81 ---------------------------------------- 3.67/1.81 3.67/1.81 (5) DependencyGraphProof (EQUIVALENT) 3.67/1.81 The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 2 less nodes. 3.67/1.81 ---------------------------------------- 3.67/1.81 3.67/1.81 (6) 3.67/1.81 Obligation: 3.67/1.81 Pi DP problem: 3.67/1.81 The TRS P consists of the following rules: 3.67/1.81 3.67/1.81 AVG_IN_GAG(X, s(s(s(Y))), s(Z)) -> AVG_IN_GAG(s(X), Y, Z) 3.67/1.81 AVG_IN_GAG(s(X), Y, Z) -> AVG_IN_GAG(X, s(Y), Z) 3.67/1.81 3.67/1.81 The TRS R consists of the following rules: 3.67/1.81 3.67/1.81 avg_in_gag(s(X), Y, Z) -> U1_gag(X, Y, Z, avg_in_gag(X, s(Y), Z)) 3.67/1.81 avg_in_gag(X, s(s(s(Y))), s(Z)) -> U2_gag(X, Y, Z, avg_in_gag(s(X), Y, Z)) 3.67/1.81 avg_in_gag(0, 0, 0) -> avg_out_gag(0, 0, 0) 3.67/1.81 avg_in_gag(0, s(0), 0) -> avg_out_gag(0, s(0), 0) 3.67/1.81 avg_in_gag(0, s(s(0)), s(0)) -> avg_out_gag(0, s(s(0)), s(0)) 3.67/1.81 U2_gag(X, Y, Z, avg_out_gag(s(X), Y, Z)) -> avg_out_gag(X, s(s(s(Y))), s(Z)) 3.67/1.81 U1_gag(X, Y, Z, avg_out_gag(X, s(Y), Z)) -> avg_out_gag(s(X), Y, Z) 3.67/1.81 3.67/1.81 The argument filtering Pi contains the following mapping: 3.67/1.81 avg_in_gag(x1, x2, x3) = avg_in_gag(x1, x3) 3.67/1.81 3.67/1.81 s(x1) = s(x1) 3.67/1.81 3.67/1.81 U1_gag(x1, x2, x3, x4) = U1_gag(x4) 3.67/1.81 3.67/1.81 U2_gag(x1, x2, x3, x4) = U2_gag(x4) 3.67/1.81 3.67/1.81 0 = 0 3.67/1.81 3.67/1.81 avg_out_gag(x1, x2, x3) = avg_out_gag(x2) 3.67/1.81 3.67/1.81 AVG_IN_GAG(x1, x2, x3) = AVG_IN_GAG(x1, x3) 3.67/1.81 3.67/1.81 3.67/1.81 We have to consider all (P,R,Pi)-chains 3.67/1.81 ---------------------------------------- 3.67/1.81 3.67/1.81 (7) UsableRulesProof (EQUIVALENT) 3.67/1.81 For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. 3.67/1.81 ---------------------------------------- 3.67/1.81 3.67/1.81 (8) 3.67/1.81 Obligation: 3.67/1.81 Pi DP problem: 3.67/1.81 The TRS P consists of the following rules: 3.67/1.81 3.67/1.81 AVG_IN_GAG(X, s(s(s(Y))), s(Z)) -> AVG_IN_GAG(s(X), Y, Z) 3.67/1.81 AVG_IN_GAG(s(X), Y, Z) -> AVG_IN_GAG(X, s(Y), Z) 3.67/1.81 3.67/1.81 R is empty. 3.67/1.81 The argument filtering Pi contains the following mapping: 3.67/1.81 s(x1) = s(x1) 3.67/1.81 3.67/1.81 AVG_IN_GAG(x1, x2, x3) = AVG_IN_GAG(x1, x3) 3.67/1.81 3.67/1.81 3.67/1.81 We have to consider all (P,R,Pi)-chains 3.67/1.81 ---------------------------------------- 3.67/1.81 3.67/1.81 (9) PiDPToQDPProof (SOUND) 3.67/1.81 Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. 3.67/1.81 ---------------------------------------- 3.67/1.81 3.67/1.81 (10) 3.67/1.81 Obligation: 3.67/1.81 Q DP problem: 3.67/1.81 The TRS P consists of the following rules: 3.67/1.81 3.67/1.81 AVG_IN_GAG(X, s(Z)) -> AVG_IN_GAG(s(X), Z) 3.67/1.81 AVG_IN_GAG(s(X), Z) -> AVG_IN_GAG(X, Z) 3.67/1.81 3.67/1.81 R is empty. 3.67/1.81 Q is empty. 3.67/1.81 We have to consider all (P,Q,R)-chains. 3.67/1.81 ---------------------------------------- 3.67/1.81 3.67/1.81 (11) QDPSizeChangeProof (EQUIVALENT) 3.67/1.81 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. 3.67/1.81 3.67/1.81 From the DPs we obtained the following set of size-change graphs: 3.67/1.81 *AVG_IN_GAG(X, s(Z)) -> AVG_IN_GAG(s(X), Z) 3.67/1.81 The graph contains the following edges 2 > 2 3.67/1.81 3.67/1.81 3.67/1.81 *AVG_IN_GAG(s(X), Z) -> AVG_IN_GAG(X, Z) 3.67/1.81 The graph contains the following edges 1 > 1, 2 >= 2 3.67/1.81 3.67/1.81 3.67/1.81 ---------------------------------------- 3.67/1.81 3.67/1.81 (12) 3.67/1.81 YES 3.85/1.85 EOF