/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given RelTRS could be proven: (0) RelTRS (1) RelTRS Reverse [EQUIVALENT, 0 ms] (2) RelTRS (3) RelTRSRRRProof [EQUIVALENT, 1250 ms] (4) RelTRS (5) RelTRSRRRProof [EQUIVALENT, 57 ms] (6) RelTRS (7) RelTRSRRRProof [EQUIVALENT, 0 ms] (8) RelTRS (9) RelTRSRRRProof [EQUIVALENT, 33 ms] (10) RelTRS (11) SIsEmptyProof [EQUIVALENT, 0 ms] (12) QTRS (13) RFCMatchBoundsTRSProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: c(a(b(x1))) -> c(b(b(x1))) b(c(c(x1))) -> c(c(c(x1))) a(a(a(x1))) -> c(c(b(x1))) The relative TRS consists of the following S rules: c(b(b(x1))) -> c(c(c(x1))) c(a(c(x1))) -> a(a(b(x1))) b(c(b(x1))) -> b(b(b(x1))) ---------------------------------------- (1) RelTRS Reverse (EQUIVALENT) We have reversed the following relative TRS [REVERSE]: The set of rules R is c(a(b(x1))) -> c(b(b(x1))) b(c(c(x1))) -> c(c(c(x1))) a(a(a(x1))) -> c(c(b(x1))) The set of rules S is c(b(b(x1))) -> c(c(c(x1))) c(a(c(x1))) -> a(a(b(x1))) b(c(b(x1))) -> b(b(b(x1))) We have obtained the following relative TRS: The set of rules R is b(a(c(x1))) -> b(b(c(x1))) c(c(b(x1))) -> c(c(c(x1))) a(a(a(x1))) -> b(c(c(x1))) The set of rules S is b(b(c(x1))) -> c(c(c(x1))) c(a(c(x1))) -> b(a(a(x1))) b(c(b(x1))) -> b(b(b(x1))) ---------------------------------------- (2) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: b(a(c(x1))) -> b(b(c(x1))) c(c(b(x1))) -> c(c(c(x1))) a(a(a(x1))) -> b(c(c(x1))) The relative TRS consists of the following S rules: b(b(c(x1))) -> c(c(c(x1))) c(a(c(x1))) -> b(a(a(x1))) b(c(b(x1))) -> b(b(b(x1))) ---------------------------------------- (3) RelTRSRRRProof (EQUIVALENT) We used the following monotonic ordering for rule removal: Matrix interpretation [MATRO] to (N^6, +, *, >=, >) : <<< POL(b(x_1)) = [[0], [0], [0], [0], [0], [0]] + [[1, 0, 1, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]] * x_1 >>> <<< POL(a(x_1)) = [[0], [0], [0], [0], [1], [0]] + [[1, 1, 0, 1, 0, 0], [0, 0, 0, 1, 0, 0], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 1, 0], [0, 0, 1, 0, 0, 0], [0, 0, 0, 0, 0, 0]] * x_1 >>> <<< POL(c(x_1)) = [[0], [0], [0], [0], [0], [0]] + [[1, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 1, 0, 1, 1, 0]] * x_1 >>> With this ordering the following rules can be removed [MATRO] because they are oriented strictly: Rules from R: a(a(a(x1))) -> b(c(c(x1))) Rules from S: none ---------------------------------------- (4) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: b(a(c(x1))) -> b(b(c(x1))) c(c(b(x1))) -> c(c(c(x1))) The relative TRS consists of the following S rules: b(b(c(x1))) -> c(c(c(x1))) c(a(c(x1))) -> b(a(a(x1))) b(c(b(x1))) -> b(b(b(x1))) ---------------------------------------- (5) RelTRSRRRProof (EQUIVALENT) We used the following monotonic ordering for rule removal: Matrix interpretation [MATRO] to (N^2, +, *, >=, >) : <<< POL(b(x_1)) = [[0], [0]] + [[2, 2], [0, 0]] * x_1 >>> <<< POL(a(x_1)) = [[0], [2]] + [[1, 0], [2, 0]] * x_1 >>> <<< POL(c(x_1)) = [[0], [0]] + [[2, 2], [0, 0]] * x_1 >>> With this ordering the following rules can be removed [MATRO] because they are oriented strictly: Rules from R: b(a(c(x1))) -> b(b(c(x1))) Rules from S: none ---------------------------------------- (6) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: c(c(b(x1))) -> c(c(c(x1))) The relative TRS consists of the following S rules: b(b(c(x1))) -> c(c(c(x1))) c(a(c(x1))) -> b(a(a(x1))) b(c(b(x1))) -> b(b(b(x1))) ---------------------------------------- (7) RelTRSRRRProof (EQUIVALENT) We used the following monotonic ordering for rule removal: Polynomial interpretation [POLO]: POL(a(x_1)) = x_1 POL(b(x_1)) = 1 + x_1 POL(c(x_1)) = 1 + x_1 With this ordering the following rules can be removed [MATRO] because they are oriented strictly: Rules from R: none Rules from S: c(a(c(x1))) -> b(a(a(x1))) ---------------------------------------- (8) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: c(c(b(x1))) -> c(c(c(x1))) The relative TRS consists of the following S rules: b(b(c(x1))) -> c(c(c(x1))) b(c(b(x1))) -> b(b(b(x1))) ---------------------------------------- (9) RelTRSRRRProof (EQUIVALENT) We used the following monotonic ordering for rule removal: Matrix interpretation [MATRO] to (N^2, +, *, >=, >) : <<< POL(c(x_1)) = [[0], [1]] + [[1, 0], [2, 0]] * x_1 >>> <<< POL(b(x_1)) = [[0], [0]] + [[1, 1], [0, 1]] * x_1 >>> With this ordering the following rules can be removed [MATRO] because they are oriented strictly: Rules from R: none Rules from S: b(b(c(x1))) -> c(c(c(x1))) b(c(b(x1))) -> b(b(b(x1))) ---------------------------------------- (10) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: c(c(b(x1))) -> c(c(c(x1))) S is empty. ---------------------------------------- (11) SIsEmptyProof (EQUIVALENT) The TRS S is empty. Hence, termination of R/S is equivalent to termination of R. ---------------------------------------- (12) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: c(c(b(x1))) -> c(c(c(x1))) Q is empty. ---------------------------------------- (13) RFCMatchBoundsTRSProof (EQUIVALENT) Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 1. This implies Q-termination of R. The following rules were used to construct the certificate: c(c(b(x1))) -> c(c(c(x1))) The certificate found is represented by the following graph. The certificate consists of the following enumerated nodes: 153, 154, 155, 156, 157, 158 Node 153 is start node and node 154 is final node. Those nodes are connected through the following edges: * 153 to 155 labelled c_1(0)* 154 to 154 labelled #_1(0)* 155 to 156 labelled c_1(0)* 155 to 157 labelled c_1(1)* 156 to 154 labelled c_1(0)* 156 to 157 labelled c_1(1)* 157 to 158 labelled c_1(1)* 157 to 157 labelled c_1(1)* 158 to 154 labelled c_1(1)* 158 to 157 labelled c_1(1) ---------------------------------------- (14) YES