/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- WORST_CASE(Omega(n^1), O(n^2)) proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^2). (0) CpxTRS (1) CpxTrsToCdtProof [UPPER BOUND(ID), 0 ms] (2) CdtProblem (3) CdtLeafRemovalProof [BOTH BOUNDS(ID, ID), 0 ms] (4) CdtProblem (5) CdtUsableRulesProof [BOTH BOUNDS(ID, ID), 0 ms] (6) CdtProblem (7) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 93 ms] (8) CdtProblem (9) CdtRuleRemovalProof [UPPER BOUND(ADD(n^2)), 169 ms] (10) CdtProblem (11) CdtRuleRemovalProof [UPPER BOUND(ADD(n^2)), 89 ms] (12) CdtProblem (13) CdtRuleRemovalProof [UPPER BOUND(ADD(n^2)), 89 ms] (14) CdtProblem (15) CdtKnowledgeProof [FINISHED, 0 ms] (16) BOUNDS(1, 1) (17) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms] (18) TRS for Loop Detection (19) DecreasingLoopProof [LOWER BOUND(ID), 0 ms] (20) BEST (21) proven lower bound (22) LowerBoundPropagationProof [FINISHED, 0 ms] (23) BOUNDS(n^1, INF) (24) TRS for Loop Detection ---------------------------------------- (0) Obligation: The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^2). The TRS R consists of the following rules: eq(0, 0) -> true eq(0, s(X)) -> false eq(s(X), 0) -> false eq(s(X), s(Y)) -> eq(X, Y) rm(N, nil) -> nil rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X)) ifrm(true, N, add(M, X)) -> rm(N, X) ifrm(false, N, add(M, X)) -> add(M, rm(N, X)) purge(nil) -> nil purge(add(N, X)) -> add(N, purge(rm(N, X))) S is empty. Rewrite Strategy: INNERMOST ---------------------------------------- (1) CpxTrsToCdtProof (UPPER BOUND(ID)) Converted Cpx (relative) TRS to CDT ---------------------------------------- (2) Obligation: Complexity Dependency Tuples Problem Rules: eq(0, 0) -> true eq(0, s(z0)) -> false eq(s(z0), 0) -> false eq(s(z0), s(z1)) -> eq(z0, z1) rm(z0, nil) -> nil rm(z0, add(z1, z2)) -> ifrm(eq(z0, z1), z0, add(z1, z2)) ifrm(true, z0, add(z1, z2)) -> rm(z0, z2) ifrm(false, z0, add(z1, z2)) -> add(z1, rm(z0, z2)) purge(nil) -> nil purge(add(z0, z1)) -> add(z0, purge(rm(z0, z1))) Tuples: EQ(0, 0) -> c EQ(0, s(z0)) -> c1 EQ(s(z0), 0) -> c2 EQ(s(z0), s(z1)) -> c3(EQ(z0, z1)) RM(z0, nil) -> c4 RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1)) IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2)) IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2)) PURGE(nil) -> c8 PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1)) S tuples: EQ(0, 0) -> c EQ(0, s(z0)) -> c1 EQ(s(z0), 0) -> c2 EQ(s(z0), s(z1)) -> c3(EQ(z0, z1)) RM(z0, nil) -> c4 RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1)) IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2)) IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2)) PURGE(nil) -> c8 PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1)) K tuples:none Defined Rule Symbols: eq_2, rm_2, ifrm_3, purge_1 Defined Pair Symbols: EQ_2, RM_2, IFRM_3, PURGE_1 Compound Symbols: c, c1, c2, c3_1, c4, c5_2, c6_1, c7_1, c8, c9_2 ---------------------------------------- (3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID)) Removed 5 trailing nodes: PURGE(nil) -> c8 EQ(s(z0), 0) -> c2 RM(z0, nil) -> c4 EQ(0, s(z0)) -> c1 EQ(0, 0) -> c ---------------------------------------- (4) Obligation: Complexity Dependency Tuples Problem Rules: eq(0, 0) -> true eq(0, s(z0)) -> false eq(s(z0), 0) -> false eq(s(z0), s(z1)) -> eq(z0, z1) rm(z0, nil) -> nil rm(z0, add(z1, z2)) -> ifrm(eq(z0, z1), z0, add(z1, z2)) ifrm(true, z0, add(z1, z2)) -> rm(z0, z2) ifrm(false, z0, add(z1, z2)) -> add(z1, rm(z0, z2)) purge(nil) -> nil purge(add(z0, z1)) -> add(z0, purge(rm(z0, z1))) Tuples: EQ(s(z0), s(z1)) -> c3(EQ(z0, z1)) RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1)) IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2)) IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2)) PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1)) S tuples: EQ(s(z0), s(z1)) -> c3(EQ(z0, z1)) RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1)) IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2)) IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2)) PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1)) K tuples:none Defined Rule Symbols: eq_2, rm_2, ifrm_3, purge_1 Defined Pair Symbols: EQ_2, RM_2, IFRM_3, PURGE_1 Compound Symbols: c3_1, c5_2, c6_1, c7_1, c9_2 ---------------------------------------- (5) CdtUsableRulesProof (BOTH BOUNDS(ID, ID)) The following rules are not usable and were removed: purge(nil) -> nil purge(add(z0, z1)) -> add(z0, purge(rm(z0, z1))) ---------------------------------------- (6) Obligation: Complexity Dependency Tuples Problem Rules: eq(0, 0) -> true eq(0, s(z0)) -> false eq(s(z0), 0) -> false eq(s(z0), s(z1)) -> eq(z0, z1) rm(z0, nil) -> nil rm(z0, add(z1, z2)) -> ifrm(eq(z0, z1), z0, add(z1, z2)) ifrm(true, z0, add(z1, z2)) -> rm(z0, z2) ifrm(false, z0, add(z1, z2)) -> add(z1, rm(z0, z2)) Tuples: EQ(s(z0), s(z1)) -> c3(EQ(z0, z1)) RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1)) IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2)) IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2)) PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1)) S tuples: EQ(s(z0), s(z1)) -> c3(EQ(z0, z1)) RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1)) IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2)) IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2)) PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1)) K tuples:none Defined Rule Symbols: eq_2, rm_2, ifrm_3 Defined Pair Symbols: EQ_2, RM_2, IFRM_3, PURGE_1 Compound Symbols: c3_1, c5_2, c6_1, c7_1, c9_2 ---------------------------------------- (7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1))) Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S. PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1)) We considered the (Usable) Rules: rm(z0, add(z1, z2)) -> ifrm(eq(z0, z1), z0, add(z1, z2)) ifrm(false, z0, add(z1, z2)) -> add(z1, rm(z0, z2)) rm(z0, nil) -> nil ifrm(true, z0, add(z1, z2)) -> rm(z0, z2) And the Tuples: EQ(s(z0), s(z1)) -> c3(EQ(z0, z1)) RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1)) IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2)) IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2)) PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1)) The order we found is given by the following interpretation: Polynomial interpretation : POL(0) = [3] POL(EQ(x_1, x_2)) = 0 POL(IFRM(x_1, x_2, x_3)) = [1] POL(PURGE(x_1)) = x_1 POL(RM(x_1, x_2)) = [1] POL(add(x_1, x_2)) = [2] + x_2 POL(c3(x_1)) = x_1 POL(c5(x_1, x_2)) = x_1 + x_2 POL(c6(x_1)) = x_1 POL(c7(x_1)) = x_1 POL(c9(x_1, x_2)) = x_1 + x_2 POL(eq(x_1, x_2)) = 0 POL(false) = 0 POL(ifrm(x_1, x_2, x_3)) = x_3 POL(nil) = 0 POL(rm(x_1, x_2)) = x_2 POL(s(x_1)) = x_1 POL(true) = 0 ---------------------------------------- (8) Obligation: Complexity Dependency Tuples Problem Rules: eq(0, 0) -> true eq(0, s(z0)) -> false eq(s(z0), 0) -> false eq(s(z0), s(z1)) -> eq(z0, z1) rm(z0, nil) -> nil rm(z0, add(z1, z2)) -> ifrm(eq(z0, z1), z0, add(z1, z2)) ifrm(true, z0, add(z1, z2)) -> rm(z0, z2) ifrm(false, z0, add(z1, z2)) -> add(z1, rm(z0, z2)) Tuples: EQ(s(z0), s(z1)) -> c3(EQ(z0, z1)) RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1)) IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2)) IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2)) PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1)) S tuples: EQ(s(z0), s(z1)) -> c3(EQ(z0, z1)) RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1)) IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2)) IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2)) K tuples: PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1)) Defined Rule Symbols: eq_2, rm_2, ifrm_3 Defined Pair Symbols: EQ_2, RM_2, IFRM_3, PURGE_1 Compound Symbols: c3_1, c5_2, c6_1, c7_1, c9_2 ---------------------------------------- (9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2))) Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S. EQ(s(z0), s(z1)) -> c3(EQ(z0, z1)) We considered the (Usable) Rules: rm(z0, add(z1, z2)) -> ifrm(eq(z0, z1), z0, add(z1, z2)) ifrm(false, z0, add(z1, z2)) -> add(z1, rm(z0, z2)) rm(z0, nil) -> nil ifrm(true, z0, add(z1, z2)) -> rm(z0, z2) And the Tuples: EQ(s(z0), s(z1)) -> c3(EQ(z0, z1)) RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1)) IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2)) IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2)) PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1)) The order we found is given by the following interpretation: Polynomial interpretation : POL(0) = 0 POL(EQ(x_1, x_2)) = x_1 POL(IFRM(x_1, x_2, x_3)) = [2]x_2*x_3 POL(PURGE(x_1)) = [2]x_1^2 POL(RM(x_1, x_2)) = [2]x_1 + [2]x_1*x_2 POL(add(x_1, x_2)) = [2] + x_1 + x_2 POL(c3(x_1)) = x_1 POL(c5(x_1, x_2)) = x_1 + x_2 POL(c6(x_1)) = x_1 POL(c7(x_1)) = x_1 POL(c9(x_1, x_2)) = x_1 + x_2 POL(eq(x_1, x_2)) = 0 POL(false) = 0 POL(ifrm(x_1, x_2, x_3)) = x_3 POL(nil) = 0 POL(rm(x_1, x_2)) = x_2 POL(s(x_1)) = [2] + x_1 POL(true) = 0 ---------------------------------------- (10) Obligation: Complexity Dependency Tuples Problem Rules: eq(0, 0) -> true eq(0, s(z0)) -> false eq(s(z0), 0) -> false eq(s(z0), s(z1)) -> eq(z0, z1) rm(z0, nil) -> nil rm(z0, add(z1, z2)) -> ifrm(eq(z0, z1), z0, add(z1, z2)) ifrm(true, z0, add(z1, z2)) -> rm(z0, z2) ifrm(false, z0, add(z1, z2)) -> add(z1, rm(z0, z2)) Tuples: EQ(s(z0), s(z1)) -> c3(EQ(z0, z1)) RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1)) IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2)) IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2)) PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1)) S tuples: RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1)) IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2)) IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2)) K tuples: PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1)) EQ(s(z0), s(z1)) -> c3(EQ(z0, z1)) Defined Rule Symbols: eq_2, rm_2, ifrm_3 Defined Pair Symbols: EQ_2, RM_2, IFRM_3, PURGE_1 Compound Symbols: c3_1, c5_2, c6_1, c7_1, c9_2 ---------------------------------------- (11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2))) Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S. IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2)) We considered the (Usable) Rules: rm(z0, add(z1, z2)) -> ifrm(eq(z0, z1), z0, add(z1, z2)) eq(0, s(z0)) -> false eq(s(z0), 0) -> false eq(s(z0), s(z1)) -> eq(z0, z1) eq(0, 0) -> true ifrm(false, z0, add(z1, z2)) -> add(z1, rm(z0, z2)) rm(z0, nil) -> nil ifrm(true, z0, add(z1, z2)) -> rm(z0, z2) And the Tuples: EQ(s(z0), s(z1)) -> c3(EQ(z0, z1)) RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1)) IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2)) IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2)) PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1)) The order we found is given by the following interpretation: Polynomial interpretation : POL(0) = [1] POL(EQ(x_1, x_2)) = 0 POL(IFRM(x_1, x_2, x_3)) = x_1 + [2]x_2*x_3 POL(PURGE(x_1)) = x_1^2 POL(RM(x_1, x_2)) = [2]x_1 + [2]x_1*x_2 POL(add(x_1, x_2)) = [1] + x_1 + x_2 POL(c3(x_1)) = x_1 POL(c5(x_1, x_2)) = x_1 + x_2 POL(c6(x_1)) = x_1 POL(c7(x_1)) = x_1 POL(c9(x_1, x_2)) = x_1 + x_2 POL(eq(x_1, x_2)) = x_1 POL(false) = [1] POL(ifrm(x_1, x_2, x_3)) = x_3 POL(nil) = 0 POL(rm(x_1, x_2)) = x_2 POL(s(x_1)) = [1] + x_1 POL(true) = 0 ---------------------------------------- (12) Obligation: Complexity Dependency Tuples Problem Rules: eq(0, 0) -> true eq(0, s(z0)) -> false eq(s(z0), 0) -> false eq(s(z0), s(z1)) -> eq(z0, z1) rm(z0, nil) -> nil rm(z0, add(z1, z2)) -> ifrm(eq(z0, z1), z0, add(z1, z2)) ifrm(true, z0, add(z1, z2)) -> rm(z0, z2) ifrm(false, z0, add(z1, z2)) -> add(z1, rm(z0, z2)) Tuples: EQ(s(z0), s(z1)) -> c3(EQ(z0, z1)) RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1)) IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2)) IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2)) PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1)) S tuples: RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1)) IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2)) K tuples: PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1)) EQ(s(z0), s(z1)) -> c3(EQ(z0, z1)) IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2)) Defined Rule Symbols: eq_2, rm_2, ifrm_3 Defined Pair Symbols: EQ_2, RM_2, IFRM_3, PURGE_1 Compound Symbols: c3_1, c5_2, c6_1, c7_1, c9_2 ---------------------------------------- (13) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2))) Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S. IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2)) We considered the (Usable) Rules: rm(z0, add(z1, z2)) -> ifrm(eq(z0, z1), z0, add(z1, z2)) ifrm(false, z0, add(z1, z2)) -> add(z1, rm(z0, z2)) rm(z0, nil) -> nil ifrm(true, z0, add(z1, z2)) -> rm(z0, z2) And the Tuples: EQ(s(z0), s(z1)) -> c3(EQ(z0, z1)) RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1)) IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2)) IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2)) PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1)) The order we found is given by the following interpretation: Polynomial interpretation : POL(0) = 0 POL(EQ(x_1, x_2)) = 0 POL(IFRM(x_1, x_2, x_3)) = [2]x_3 POL(PURGE(x_1)) = [2]x_1^2 POL(RM(x_1, x_2)) = [2]x_2 POL(add(x_1, x_2)) = [1] + x_2 POL(c3(x_1)) = x_1 POL(c5(x_1, x_2)) = x_1 + x_2 POL(c6(x_1)) = x_1 POL(c7(x_1)) = x_1 POL(c9(x_1, x_2)) = x_1 + x_2 POL(eq(x_1, x_2)) = 0 POL(false) = 0 POL(ifrm(x_1, x_2, x_3)) = x_3 POL(nil) = 0 POL(rm(x_1, x_2)) = x_2 POL(s(x_1)) = 0 POL(true) = 0 ---------------------------------------- (14) Obligation: Complexity Dependency Tuples Problem Rules: eq(0, 0) -> true eq(0, s(z0)) -> false eq(s(z0), 0) -> false eq(s(z0), s(z1)) -> eq(z0, z1) rm(z0, nil) -> nil rm(z0, add(z1, z2)) -> ifrm(eq(z0, z1), z0, add(z1, z2)) ifrm(true, z0, add(z1, z2)) -> rm(z0, z2) ifrm(false, z0, add(z1, z2)) -> add(z1, rm(z0, z2)) Tuples: EQ(s(z0), s(z1)) -> c3(EQ(z0, z1)) RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1)) IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2)) IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2)) PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1)) S tuples: RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1)) K tuples: PURGE(add(z0, z1)) -> c9(PURGE(rm(z0, z1)), RM(z0, z1)) EQ(s(z0), s(z1)) -> c3(EQ(z0, z1)) IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2)) IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2)) Defined Rule Symbols: eq_2, rm_2, ifrm_3 Defined Pair Symbols: EQ_2, RM_2, IFRM_3, PURGE_1 Compound Symbols: c3_1, c5_2, c6_1, c7_1, c9_2 ---------------------------------------- (15) CdtKnowledgeProof (FINISHED) The following tuples could be moved from S to K by knowledge propagation: RM(z0, add(z1, z2)) -> c5(IFRM(eq(z0, z1), z0, add(z1, z2)), EQ(z0, z1)) IFRM(true, z0, add(z1, z2)) -> c6(RM(z0, z2)) IFRM(false, z0, add(z1, z2)) -> c7(RM(z0, z2)) EQ(s(z0), s(z1)) -> c3(EQ(z0, z1)) Now S is empty ---------------------------------------- (16) BOUNDS(1, 1) ---------------------------------------- (17) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID)) Transformed a relative TRS into a decreasing-loop problem. ---------------------------------------- (18) Obligation: Analyzing the following TRS for decreasing loops: The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^2). The TRS R consists of the following rules: eq(0, 0) -> true eq(0, s(X)) -> false eq(s(X), 0) -> false eq(s(X), s(Y)) -> eq(X, Y) rm(N, nil) -> nil rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X)) ifrm(true, N, add(M, X)) -> rm(N, X) ifrm(false, N, add(M, X)) -> add(M, rm(N, X)) purge(nil) -> nil purge(add(N, X)) -> add(N, purge(rm(N, X))) S is empty. Rewrite Strategy: INNERMOST ---------------------------------------- (19) DecreasingLoopProof (LOWER BOUND(ID)) The following loop(s) give(s) rise to the lower bound Omega(n^1): The rewrite sequence eq(s(X), s(Y)) ->^+ eq(X, Y) gives rise to a decreasing loop by considering the right hand sides subterm at position []. The pumping substitution is [X / s(X), Y / s(Y)]. The result substitution is [ ]. ---------------------------------------- (20) Complex Obligation (BEST) ---------------------------------------- (21) Obligation: Proved the lower bound n^1 for the following obligation: The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^2). The TRS R consists of the following rules: eq(0, 0) -> true eq(0, s(X)) -> false eq(s(X), 0) -> false eq(s(X), s(Y)) -> eq(X, Y) rm(N, nil) -> nil rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X)) ifrm(true, N, add(M, X)) -> rm(N, X) ifrm(false, N, add(M, X)) -> add(M, rm(N, X)) purge(nil) -> nil purge(add(N, X)) -> add(N, purge(rm(N, X))) S is empty. Rewrite Strategy: INNERMOST ---------------------------------------- (22) LowerBoundPropagationProof (FINISHED) Propagated lower bound. ---------------------------------------- (23) BOUNDS(n^1, INF) ---------------------------------------- (24) Obligation: Analyzing the following TRS for decreasing loops: The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^2). The TRS R consists of the following rules: eq(0, 0) -> true eq(0, s(X)) -> false eq(s(X), 0) -> false eq(s(X), s(Y)) -> eq(X, Y) rm(N, nil) -> nil rm(N, add(M, X)) -> ifrm(eq(N, M), N, add(M, X)) ifrm(true, N, add(M, X)) -> rm(N, X) ifrm(false, N, add(M, X)) -> add(M, rm(N, X)) purge(nil) -> nil purge(add(N, X)) -> add(N, purge(rm(N, X))) S is empty. Rewrite Strategy: INNERMOST