3.72/1.91 YES 3.72/1.92 proof of /export/starexec/sandbox/benchmark/theBenchmark.itrs 3.72/1.92 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.72/1.92 3.72/1.92 3.72/1.92 Termination of the given ITRS could be proven: 3.72/1.92 3.72/1.92 (0) ITRS 3.72/1.92 (1) ITRStoIDPProof [EQUIVALENT, 0 ms] 3.72/1.92 (2) IDP 3.72/1.92 (3) UsableRulesProof [EQUIVALENT, 0 ms] 3.72/1.92 (4) IDP 3.72/1.92 (5) IDependencyGraphProof [EQUIVALENT, 0 ms] 3.72/1.92 (6) IDP 3.72/1.92 (7) IDPNonInfProof [SOUND, 171 ms] 3.72/1.92 (8) IDP 3.72/1.92 (9) PisEmptyProof [EQUIVALENT, 0 ms] 3.72/1.92 (10) YES 3.72/1.92 3.72/1.92 3.72/1.92 ---------------------------------------- 3.72/1.92 3.72/1.92 (0) 3.72/1.92 Obligation: 3.72/1.92 ITRS problem: 3.72/1.92 3.72/1.92 The following function symbols are pre-defined: 3.72/1.92 <<< 3.72/1.92 & ~ Bwand: (Integer, Integer) -> Integer 3.72/1.92 >= ~ Ge: (Integer, Integer) -> Boolean 3.72/1.92 | ~ Bwor: (Integer, Integer) -> Integer 3.72/1.92 / ~ Div: (Integer, Integer) -> Integer 3.72/1.92 != ~ Neq: (Integer, Integer) -> Boolean 3.72/1.92 && ~ Land: (Boolean, Boolean) -> Boolean 3.72/1.92 ! ~ Lnot: (Boolean) -> Boolean 3.72/1.92 = ~ Eq: (Integer, Integer) -> Boolean 3.72/1.92 <= ~ Le: (Integer, Integer) -> Boolean 3.72/1.92 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.72/1.92 % ~ Mod: (Integer, Integer) -> Integer 3.72/1.92 + ~ Add: (Integer, Integer) -> Integer 3.72/1.92 > ~ Gt: (Integer, Integer) -> Boolean 3.72/1.92 -1 ~ UnaryMinus: (Integer) -> Integer 3.72/1.92 < ~ Lt: (Integer, Integer) -> Boolean 3.72/1.92 || ~ Lor: (Boolean, Boolean) -> Boolean 3.72/1.92 - ~ Sub: (Integer, Integer) -> Integer 3.72/1.92 ~ ~ Bwnot: (Integer) -> Integer 3.72/1.92 * ~ Mul: (Integer, Integer) -> Integer 3.72/1.92 >>> 3.72/1.92 3.72/1.92 The TRS R consists of the following rules: 3.72/1.92 f(TRUE, x) -> ft(TRUE, x, y) 3.72/1.92 ft(TRUE, x, y) -> ft(y >= x, x + 1, y) 3.72/1.92 The set Q consists of the following terms: 3.72/1.92 f(TRUE, x0) 3.72/1.92 ft(TRUE, x0, x1) 3.72/1.92 3.72/1.92 ---------------------------------------- 3.72/1.92 3.72/1.92 (1) ITRStoIDPProof (EQUIVALENT) 3.72/1.92 Added dependency pairs 3.72/1.92 ---------------------------------------- 3.72/1.92 3.72/1.92 (2) 3.72/1.92 Obligation: 3.72/1.92 IDP problem: 3.72/1.92 The following function symbols are pre-defined: 3.72/1.92 <<< 3.72/1.92 & ~ Bwand: (Integer, Integer) -> Integer 3.72/1.92 >= ~ Ge: (Integer, Integer) -> Boolean 3.72/1.92 | ~ Bwor: (Integer, Integer) -> Integer 3.72/1.92 / ~ Div: (Integer, Integer) -> Integer 3.72/1.92 != ~ Neq: (Integer, Integer) -> Boolean 3.72/1.92 && ~ Land: (Boolean, Boolean) -> Boolean 3.72/1.92 ! ~ Lnot: (Boolean) -> Boolean 3.72/1.92 = ~ Eq: (Integer, Integer) -> Boolean 3.72/1.92 <= ~ Le: (Integer, Integer) -> Boolean 3.72/1.92 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.72/1.92 % ~ Mod: (Integer, Integer) -> Integer 3.72/1.92 + ~ Add: (Integer, Integer) -> Integer 3.72/1.92 > ~ Gt: (Integer, Integer) -> Boolean 3.72/1.92 -1 ~ UnaryMinus: (Integer) -> Integer 3.72/1.92 < ~ Lt: (Integer, Integer) -> Boolean 3.72/1.92 || ~ Lor: (Boolean, Boolean) -> Boolean 3.72/1.92 - ~ Sub: (Integer, Integer) -> Integer 3.72/1.92 ~ ~ Bwnot: (Integer) -> Integer 3.72/1.92 * ~ Mul: (Integer, Integer) -> Integer 3.72/1.92 >>> 3.72/1.92 3.72/1.92 3.72/1.92 The following domains are used: 3.72/1.92 Integer 3.72/1.92 3.72/1.92 The ITRS R consists of the following rules: 3.72/1.92 f(TRUE, x) -> ft(TRUE, x, y) 3.72/1.92 ft(TRUE, x, y) -> ft(y >= x, x + 1, y) 3.72/1.92 3.72/1.92 The integer pair graph contains the following rules and edges: 3.72/1.92 (0): F(TRUE, x[0]) -> FT(TRUE, x[0], y[0]) 3.72/1.92 (1): FT(TRUE, x[1], y[1]) -> FT(y[1] >= x[1], x[1] + 1, y[1]) 3.72/1.92 3.72/1.92 (0) -> (1), if (x[0] ->^* x[1] & y[0] ->^* y[1]) 3.72/1.92 (1) -> (1), if (y[1] >= x[1] & x[1] + 1 ->^* x[1]' & y[1] ->^* y[1]') 3.72/1.92 3.72/1.92 The set Q consists of the following terms: 3.72/1.92 f(TRUE, x0) 3.72/1.92 ft(TRUE, x0, x1) 3.72/1.92 3.72/1.92 ---------------------------------------- 3.72/1.92 3.72/1.92 (3) UsableRulesProof (EQUIVALENT) 3.72/1.92 As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. 3.72/1.92 ---------------------------------------- 3.72/1.92 3.72/1.92 (4) 3.72/1.92 Obligation: 3.72/1.92 IDP problem: 3.72/1.92 The following function symbols are pre-defined: 3.72/1.92 <<< 3.72/1.92 & ~ Bwand: (Integer, Integer) -> Integer 3.72/1.92 >= ~ Ge: (Integer, Integer) -> Boolean 3.72/1.92 | ~ Bwor: (Integer, Integer) -> Integer 3.72/1.92 / ~ Div: (Integer, Integer) -> Integer 3.72/1.92 != ~ Neq: (Integer, Integer) -> Boolean 3.72/1.92 && ~ Land: (Boolean, Boolean) -> Boolean 3.72/1.92 ! ~ Lnot: (Boolean) -> Boolean 3.72/1.92 = ~ Eq: (Integer, Integer) -> Boolean 3.72/1.92 <= ~ Le: (Integer, Integer) -> Boolean 3.72/1.92 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.72/1.92 % ~ Mod: (Integer, Integer) -> Integer 3.72/1.92 + ~ Add: (Integer, Integer) -> Integer 3.72/1.92 > ~ Gt: (Integer, Integer) -> Boolean 3.72/1.92 -1 ~ UnaryMinus: (Integer) -> Integer 3.72/1.92 < ~ Lt: (Integer, Integer) -> Boolean 3.72/1.92 || ~ Lor: (Boolean, Boolean) -> Boolean 3.72/1.92 - ~ Sub: (Integer, Integer) -> Integer 3.72/1.92 ~ ~ Bwnot: (Integer) -> Integer 3.72/1.92 * ~ Mul: (Integer, Integer) -> Integer 3.72/1.92 >>> 3.72/1.92 3.72/1.92 3.72/1.92 The following domains are used: 3.72/1.92 Integer 3.72/1.92 3.72/1.92 R is empty. 3.72/1.92 3.72/1.92 The integer pair graph contains the following rules and edges: 3.72/1.92 (0): F(TRUE, x[0]) -> FT(TRUE, x[0], y[0]) 3.72/1.92 (1): FT(TRUE, x[1], y[1]) -> FT(y[1] >= x[1], x[1] + 1, y[1]) 3.72/1.92 3.72/1.92 (0) -> (1), if (x[0] ->^* x[1] & y[0] ->^* y[1]) 3.72/1.92 (1) -> (1), if (y[1] >= x[1] & x[1] + 1 ->^* x[1]' & y[1] ->^* y[1]') 3.72/1.92 3.72/1.92 The set Q consists of the following terms: 3.72/1.92 f(TRUE, x0) 3.72/1.92 ft(TRUE, x0, x1) 3.72/1.92 3.72/1.92 ---------------------------------------- 3.72/1.92 3.72/1.92 (5) IDependencyGraphProof (EQUIVALENT) 3.72/1.92 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. 3.72/1.92 ---------------------------------------- 3.72/1.92 3.72/1.92 (6) 3.72/1.92 Obligation: 3.72/1.92 IDP problem: 3.72/1.92 The following function symbols are pre-defined: 3.72/1.92 <<< 3.72/1.92 & ~ Bwand: (Integer, Integer) -> Integer 3.72/1.92 >= ~ Ge: (Integer, Integer) -> Boolean 3.72/1.92 | ~ Bwor: (Integer, Integer) -> Integer 3.72/1.92 / ~ Div: (Integer, Integer) -> Integer 3.72/1.92 != ~ Neq: (Integer, Integer) -> Boolean 3.72/1.92 && ~ Land: (Boolean, Boolean) -> Boolean 3.72/1.92 ! ~ Lnot: (Boolean) -> Boolean 3.72/1.92 = ~ Eq: (Integer, Integer) -> Boolean 3.72/1.92 <= ~ Le: (Integer, Integer) -> Boolean 3.72/1.92 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.72/1.92 % ~ Mod: (Integer, Integer) -> Integer 3.72/1.92 + ~ Add: (Integer, Integer) -> Integer 3.72/1.92 > ~ Gt: (Integer, Integer) -> Boolean 3.72/1.92 -1 ~ UnaryMinus: (Integer) -> Integer 3.72/1.92 < ~ Lt: (Integer, Integer) -> Boolean 3.72/1.92 || ~ Lor: (Boolean, Boolean) -> Boolean 3.72/1.92 - ~ Sub: (Integer, Integer) -> Integer 3.72/1.92 ~ ~ Bwnot: (Integer) -> Integer 3.72/1.92 * ~ Mul: (Integer, Integer) -> Integer 3.72/1.92 >>> 3.72/1.92 3.72/1.92 3.72/1.92 The following domains are used: 3.72/1.92 Integer 3.72/1.92 3.72/1.92 R is empty. 3.72/1.92 3.72/1.92 The integer pair graph contains the following rules and edges: 3.72/1.92 (1): FT(TRUE, x[1], y[1]) -> FT(y[1] >= x[1], x[1] + 1, y[1]) 3.72/1.92 3.72/1.92 (1) -> (1), if (y[1] >= x[1] & x[1] + 1 ->^* x[1]' & y[1] ->^* y[1]') 3.72/1.92 3.72/1.92 The set Q consists of the following terms: 3.72/1.92 f(TRUE, x0) 3.72/1.92 ft(TRUE, x0, x1) 3.72/1.92 3.72/1.92 ---------------------------------------- 3.72/1.92 3.72/1.92 (7) IDPNonInfProof (SOUND) 3.72/1.92 Used the following options for this NonInfProof: 3.72/1.92 3.72/1.92 IDPGPoloSolver: 3.72/1.92 Range: [(-1,2)] 3.72/1.92 IsNat: false 3.72/1.92 Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@7bf91d93 3.72/1.92 Constraint Generator: NonInfConstraintGenerator: 3.72/1.92 PathGenerator: MetricPathGenerator: 3.72/1.92 Max Left Steps: 1 3.72/1.92 Max Right Steps: 1 3.72/1.92 3.72/1.92 3.72/1.92 3.72/1.92 The constraints were generated the following way: 3.72/1.92 3.72/1.92 The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps: 3.72/1.92 3.72/1.92 Note that final constraints are written in bold face. 3.72/1.92 3.72/1.92 3.72/1.92 3.72/1.92 For Pair FT(TRUE, x[1], y[1]) -> FT(>=(y[1], x[1]), +(x[1], 1), y[1]) the following chains were created: 3.72/1.92 *We consider the chain FT(TRUE, x[1], y[1]) -> FT(>=(y[1], x[1]), +(x[1], 1), y[1]), FT(TRUE, x[1], y[1]) -> FT(>=(y[1], x[1]), +(x[1], 1), y[1]), FT(TRUE, x[1], y[1]) -> FT(>=(y[1], x[1]), +(x[1], 1), y[1]) which results in the following constraint: 3.72/1.92 3.72/1.92 (1) (>=(y[1], x[1])=TRUE & +(x[1], 1)=x[1]1 & y[1]=y[1]1 & >=(y[1]1, x[1]1)=TRUE & +(x[1]1, 1)=x[1]2 & y[1]1=y[1]2 ==> FT(TRUE, x[1]1, y[1]1)_>=_NonInfC & FT(TRUE, x[1]1, y[1]1)_>=_FT(>=(y[1]1, x[1]1), +(x[1]1, 1), y[1]1) & (U^Increasing(FT(>=(y[1]1, x[1]1), +(x[1]1, 1), y[1]1)), >=)) 3.72/1.92 3.72/1.92 3.72/1.92 3.72/1.92 We simplified constraint (1) using rules (III), (IV) which results in the following new constraint: 3.72/1.92 3.72/1.92 (2) (>=(y[1], x[1])=TRUE & >=(y[1], +(x[1], 1))=TRUE ==> FT(TRUE, +(x[1], 1), y[1])_>=_NonInfC & FT(TRUE, +(x[1], 1), y[1])_>=_FT(>=(y[1], +(x[1], 1)), +(+(x[1], 1), 1), y[1]) & (U^Increasing(FT(>=(y[1]1, x[1]1), +(x[1]1, 1), y[1]1)), >=)) 3.72/1.92 3.72/1.92 3.72/1.92 3.72/1.92 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 3.72/1.92 3.72/1.92 (3) (y[1] + [-1]x[1] >= 0 & y[1] + [-1] + [-1]x[1] >= 0 ==> (U^Increasing(FT(>=(y[1]1, x[1]1), +(x[1]1, 1), y[1]1)), >=) & [bni_8 + (-1)Bound*bni_8] + [bni_8]y[1] + [(-1)bni_8]x[1] >= 0 & [1 + (-1)bso_9] >= 0) 3.72/1.92 3.72/1.92 3.72/1.92 3.72/1.92 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 3.72/1.92 3.72/1.92 (4) (y[1] + [-1]x[1] >= 0 & y[1] + [-1] + [-1]x[1] >= 0 ==> (U^Increasing(FT(>=(y[1]1, x[1]1), +(x[1]1, 1), y[1]1)), >=) & [bni_8 + (-1)Bound*bni_8] + [bni_8]y[1] + [(-1)bni_8]x[1] >= 0 & [1 + (-1)bso_9] >= 0) 3.72/1.92 3.72/1.92 3.72/1.92 3.72/1.92 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 3.72/1.92 3.72/1.92 (5) (y[1] + [-1]x[1] >= 0 & y[1] + [-1] + [-1]x[1] >= 0 ==> (U^Increasing(FT(>=(y[1]1, x[1]1), +(x[1]1, 1), y[1]1)), >=) & [bni_8 + (-1)Bound*bni_8] + [bni_8]y[1] + [(-1)bni_8]x[1] >= 0 & [1 + (-1)bso_9] >= 0) 3.72/1.92 3.72/1.92 3.72/1.92 3.72/1.92 We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint: 3.72/1.92 3.72/1.92 (6) (y[1] >= 0 & [-1] + y[1] >= 0 ==> (U^Increasing(FT(>=(y[1]1, x[1]1), +(x[1]1, 1), y[1]1)), >=) & [bni_8 + (-1)Bound*bni_8] + [bni_8]y[1] >= 0 & [1 + (-1)bso_9] >= 0) 3.72/1.92 3.72/1.92 3.72/1.92 3.72/1.92 We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraints: 3.72/1.92 3.72/1.92 (7) (y[1] >= 0 & [-1] + y[1] >= 0 & x[1] >= 0 ==> (U^Increasing(FT(>=(y[1]1, x[1]1), +(x[1]1, 1), y[1]1)), >=) & [bni_8 + (-1)Bound*bni_8] + [bni_8]y[1] >= 0 & [1 + (-1)bso_9] >= 0) 3.72/1.92 3.72/1.92 (8) (y[1] >= 0 & [-1] + y[1] >= 0 & x[1] >= 0 ==> (U^Increasing(FT(>=(y[1]1, x[1]1), +(x[1]1, 1), y[1]1)), >=) & [bni_8 + (-1)Bound*bni_8] + [bni_8]y[1] >= 0 & [1 + (-1)bso_9] >= 0) 3.72/1.92 3.72/1.92 3.72/1.92 3.72/1.92 3.72/1.92 3.72/1.92 3.72/1.92 3.72/1.92 3.72/1.92 To summarize, we get the following constraints P__>=_ for the following pairs. 3.72/1.92 3.72/1.92 *FT(TRUE, x[1], y[1]) -> FT(>=(y[1], x[1]), +(x[1], 1), y[1]) 3.72/1.92 3.72/1.92 *(y[1] >= 0 & [-1] + y[1] >= 0 & x[1] >= 0 ==> (U^Increasing(FT(>=(y[1]1, x[1]1), +(x[1]1, 1), y[1]1)), >=) & [bni_8 + (-1)Bound*bni_8] + [bni_8]y[1] >= 0 & [1 + (-1)bso_9] >= 0) 3.72/1.92 3.72/1.92 3.72/1.92 *(y[1] >= 0 & [-1] + y[1] >= 0 & x[1] >= 0 ==> (U^Increasing(FT(>=(y[1]1, x[1]1), +(x[1]1, 1), y[1]1)), >=) & [bni_8 + (-1)Bound*bni_8] + [bni_8]y[1] >= 0 & [1 + (-1)bso_9] >= 0) 3.72/1.92 3.72/1.92 3.72/1.92 3.72/1.92 3.72/1.92 3.72/1.92 3.72/1.92 3.72/1.92 3.72/1.92 The constraints for P_> respective P_bound are constructed from P__>=_ where we just replace every occurence of "t _>=_ s" in P__>=_ by "t > s" respective "t _>=_ c". Here c stands for the fresh constant used for P_bound. 3.72/1.92 3.72/1.92 Using the following integer polynomial ordering the resulting constraints can be solved 3.72/1.92 3.72/1.92 Polynomial interpretation over integers[POLO]: 3.72/1.92 3.72/1.92 POL(TRUE) = 0 3.72/1.92 POL(FALSE) = 0 3.72/1.92 POL(FT(x_1, x_2, x_3)) = [2] + x_3 + [-1]x_2 3.72/1.92 POL(>=(x_1, x_2)) = [-1] 3.72/1.92 POL(+(x_1, x_2)) = x_1 + x_2 3.72/1.92 POL(1) = [1] 3.72/1.92 3.72/1.92 3.72/1.92 The following pairs are in P_>: 3.72/1.92 3.72/1.92 3.72/1.92 FT(TRUE, x[1], y[1]) -> FT(>=(y[1], x[1]), +(x[1], 1), y[1]) 3.72/1.92 3.72/1.92 3.72/1.92 The following pairs are in P_bound: 3.72/1.92 3.72/1.92 3.72/1.92 FT(TRUE, x[1], y[1]) -> FT(>=(y[1], x[1]), +(x[1], 1), y[1]) 3.72/1.92 3.72/1.92 3.72/1.92 The following pairs are in P_>=: 3.72/1.92 3.72/1.92 none 3.72/1.92 3.72/1.92 3.72/1.92 There are no usable rules. 3.72/1.92 ---------------------------------------- 3.72/1.92 3.72/1.92 (8) 3.72/1.92 Obligation: 3.72/1.92 IDP problem: 3.72/1.92 The following function symbols are pre-defined: 3.72/1.92 <<< 3.72/1.92 & ~ Bwand: (Integer, Integer) -> Integer 3.72/1.92 >= ~ Ge: (Integer, Integer) -> Boolean 3.72/1.92 | ~ Bwor: (Integer, Integer) -> Integer 3.72/1.92 / ~ Div: (Integer, Integer) -> Integer 3.72/1.92 != ~ Neq: (Integer, Integer) -> Boolean 3.72/1.92 && ~ Land: (Boolean, Boolean) -> Boolean 3.72/1.92 ! ~ Lnot: (Boolean) -> Boolean 3.72/1.92 = ~ Eq: (Integer, Integer) -> Boolean 3.72/1.92 <= ~ Le: (Integer, Integer) -> Boolean 3.72/1.92 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.72/1.92 % ~ Mod: (Integer, Integer) -> Integer 3.72/1.92 + ~ Add: (Integer, Integer) -> Integer 3.72/1.92 > ~ Gt: (Integer, Integer) -> Boolean 3.72/1.92 -1 ~ UnaryMinus: (Integer) -> Integer 3.72/1.92 < ~ Lt: (Integer, Integer) -> Boolean 3.72/1.92 || ~ Lor: (Boolean, Boolean) -> Boolean 3.72/1.92 - ~ Sub: (Integer, Integer) -> Integer 3.72/1.92 ~ ~ Bwnot: (Integer) -> Integer 3.72/1.92 * ~ Mul: (Integer, Integer) -> Integer 3.72/1.92 >>> 3.72/1.92 3.72/1.92 3.72/1.92 The following domains are used: 3.72/1.92 none 3.72/1.92 3.72/1.92 R is empty. 3.72/1.92 3.72/1.92 The integer pair graph is empty. 3.72/1.92 3.72/1.92 The set Q consists of the following terms: 3.72/1.92 f(TRUE, x0) 3.72/1.92 ft(TRUE, x0, x1) 3.72/1.92 3.72/1.92 ---------------------------------------- 3.72/1.92 3.72/1.92 (9) PisEmptyProof (EQUIVALENT) 3.72/1.92 The TRS P is empty. Hence, there is no (P,Q,R) chain. 3.72/1.92 ---------------------------------------- 3.72/1.92 3.72/1.92 (10) 3.72/1.92 YES 3.72/1.94 EOF