YES proof of prog.inttrs # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given IRSwT could be proven: (0) IRSwT (1) IRSFormatTransformerProof [EQUIVALENT, 0 ms] (2) IRSwT (3) IRSwTTerminationDigraphProof [EQUIVALENT, 79 ms] (4) IRSwT (5) IntTRSCompressionProof [EQUIVALENT, 15 ms] (6) IRSwT (7) IntTRSUnneededArgumentFilterProof [EQUIVALENT, 0 ms] (8) IRSwT (9) TempFilterProof [SOUND, 14 ms] (10) IntTRS (11) PolynomialOrderProcessor [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Rules: l0(pHAT0, yHAT0) -> l1(pHATpost, yHATpost) :|: yHAT0 = yHATpost && pHATpost = 1 && yHAT0 <= 0 l0(x, x1) -> l2(x2, x3) :|: x = x2 && x3 = -1 + x1 && 1 <= x1 l2(x4, x5) -> l0(x6, x7) :|: x5 = x7 && x4 = x6 l3(x8, x9) -> l2(x10, x11) :|: x9 = x11 && x10 = 0 l4(x12, x13) -> l3(x14, x15) :|: x13 = x15 && x12 = x14 Start term: l4(pHAT0, yHAT0) ---------------------------------------- (1) IRSFormatTransformerProof (EQUIVALENT) Reformatted IRS to match normalized format (transformed away non-linear left-hand sides, !=, / and %). ---------------------------------------- (2) Obligation: Rules: l0(pHAT0, yHAT0) -> l1(pHATpost, yHATpost) :|: yHAT0 = yHATpost && pHATpost = 1 && yHAT0 <= 0 l0(x, x1) -> l2(x2, x3) :|: x = x2 && x3 = -1 + x1 && 1 <= x1 l2(x4, x5) -> l0(x6, x7) :|: x5 = x7 && x4 = x6 l3(x8, x9) -> l2(x10, x11) :|: x9 = x11 && x10 = 0 l4(x12, x13) -> l3(x14, x15) :|: x13 = x15 && x12 = x14 Start term: l4(pHAT0, yHAT0) ---------------------------------------- (3) IRSwTTerminationDigraphProof (EQUIVALENT) Constructed termination digraph! Nodes: (1) l0(pHAT0, yHAT0) -> l1(pHATpost, yHATpost) :|: yHAT0 = yHATpost && pHATpost = 1 && yHAT0 <= 0 (2) l0(x, x1) -> l2(x2, x3) :|: x = x2 && x3 = -1 + x1 && 1 <= x1 (3) l2(x4, x5) -> l0(x6, x7) :|: x5 = x7 && x4 = x6 (4) l3(x8, x9) -> l2(x10, x11) :|: x9 = x11 && x10 = 0 (5) l4(x12, x13) -> l3(x14, x15) :|: x13 = x15 && x12 = x14 Arcs: (2) -> (3) (3) -> (1), (2) (4) -> (3) (5) -> (4) This digraph is fully evaluated! ---------------------------------------- (4) Obligation: Termination digraph: Nodes: (1) l0(x, x1) -> l2(x2, x3) :|: x = x2 && x3 = -1 + x1 && 1 <= x1 (2) l2(x4, x5) -> l0(x6, x7) :|: x5 = x7 && x4 = x6 Arcs: (1) -> (2) (2) -> (1) This digraph is fully evaluated! ---------------------------------------- (5) IntTRSCompressionProof (EQUIVALENT) Compressed rules. ---------------------------------------- (6) Obligation: Rules: l0(x2:0, x1:0) -> l0(x2:0, -1 + x1:0) :|: x1:0 > 0 ---------------------------------------- (7) IntTRSUnneededArgumentFilterProof (EQUIVALENT) Some arguments are removed because they cannot influence termination. We removed arguments according to the following replacements: l0(x1, x2) -> l0(x2) ---------------------------------------- (8) Obligation: Rules: l0(x1:0) -> l0(-1 + x1:0) :|: x1:0 > 0 ---------------------------------------- (9) TempFilterProof (SOUND) Used the following sort dictionary for filtering: l0(INTEGER) Replaced non-predefined constructor symbols by 0. ---------------------------------------- (10) Obligation: Rules: l0(x1:0) -> l0(c) :|: c = -1 + x1:0 && x1:0 > 0 ---------------------------------------- (11) PolynomialOrderProcessor (EQUIVALENT) Found the following polynomial interpretation: [l0(x)] = x The following rules are decreasing: l0(x1:0) -> l0(c) :|: c = -1 + x1:0 && x1:0 > 0 The following rules are bounded: l0(x1:0) -> l0(c) :|: c = -1 + x1:0 && x1:0 > 0 ---------------------------------------- (12) YES