3.74/1.83 YES 3.74/1.84 proof of /export/starexec/sandbox/benchmark/theBenchmark.itrs 3.74/1.84 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.74/1.84 3.74/1.84 3.74/1.84 Termination of the given ITRS could be proven: 3.74/1.84 3.74/1.84 (0) ITRS 3.74/1.84 (1) ITRStoIDPProof [EQUIVALENT, 0 ms] 3.74/1.84 (2) IDP 3.74/1.84 (3) UsableRulesProof [EQUIVALENT, 0 ms] 3.74/1.84 (4) IDP 3.74/1.84 (5) IDPNonInfProof [SOUND, 156 ms] 3.74/1.84 (6) IDP 3.74/1.84 (7) IDependencyGraphProof [EQUIVALENT, 0 ms] 3.74/1.84 (8) TRUE 3.74/1.84 3.74/1.84 3.74/1.84 ---------------------------------------- 3.74/1.84 3.74/1.84 (0) 3.74/1.84 Obligation: 3.74/1.84 ITRS problem: 3.74/1.84 3.74/1.84 The following function symbols are pre-defined: 3.74/1.84 <<< 3.74/1.84 & ~ Bwand: (Integer, Integer) -> Integer 3.74/1.84 >= ~ Ge: (Integer, Integer) -> Boolean 3.74/1.84 | ~ Bwor: (Integer, Integer) -> Integer 3.74/1.84 / ~ Div: (Integer, Integer) -> Integer 3.74/1.84 != ~ Neq: (Integer, Integer) -> Boolean 3.74/1.84 && ~ Land: (Boolean, Boolean) -> Boolean 3.74/1.84 ! ~ Lnot: (Boolean) -> Boolean 3.74/1.84 = ~ Eq: (Integer, Integer) -> Boolean 3.74/1.84 <= ~ Le: (Integer, Integer) -> Boolean 3.74/1.84 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.74/1.84 % ~ Mod: (Integer, Integer) -> Integer 3.74/1.84 > ~ Gt: (Integer, Integer) -> Boolean 3.74/1.84 + ~ Add: (Integer, Integer) -> Integer 3.74/1.84 -1 ~ UnaryMinus: (Integer) -> Integer 3.74/1.84 < ~ Lt: (Integer, Integer) -> Boolean 3.74/1.84 || ~ Lor: (Boolean, Boolean) -> Boolean 3.74/1.84 - ~ Sub: (Integer, Integer) -> Integer 3.74/1.84 ~ ~ Bwnot: (Integer) -> Integer 3.74/1.84 * ~ Mul: (Integer, Integer) -> Integer 3.74/1.84 >>> 3.74/1.84 3.74/1.84 The TRS R consists of the following rules: 3.74/1.84 div(x, y) -> if(x >= y && y > 0, x, y) 3.74/1.84 if(TRUE, x, y) -> div(x - y, y) + 1 3.74/1.84 The set Q consists of the following terms: 3.74/1.84 div(x0, x1) 3.74/1.84 if(TRUE, x0, x1) 3.74/1.84 3.74/1.84 ---------------------------------------- 3.74/1.84 3.74/1.84 (1) ITRStoIDPProof (EQUIVALENT) 3.74/1.84 Added dependency pairs 3.74/1.84 ---------------------------------------- 3.74/1.84 3.74/1.84 (2) 3.74/1.84 Obligation: 3.74/1.84 IDP problem: 3.74/1.84 The following function symbols are pre-defined: 3.74/1.84 <<< 3.74/1.84 & ~ Bwand: (Integer, Integer) -> Integer 3.74/1.84 >= ~ Ge: (Integer, Integer) -> Boolean 3.74/1.84 | ~ Bwor: (Integer, Integer) -> Integer 3.74/1.84 / ~ Div: (Integer, Integer) -> Integer 3.74/1.84 != ~ Neq: (Integer, Integer) -> Boolean 3.74/1.84 && ~ Land: (Boolean, Boolean) -> Boolean 3.74/1.84 ! ~ Lnot: (Boolean) -> Boolean 3.74/1.84 = ~ Eq: (Integer, Integer) -> Boolean 3.74/1.84 <= ~ Le: (Integer, Integer) -> Boolean 3.74/1.84 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.74/1.84 % ~ Mod: (Integer, Integer) -> Integer 3.74/1.84 > ~ Gt: (Integer, Integer) -> Boolean 3.74/1.84 + ~ Add: (Integer, Integer) -> Integer 3.74/1.84 -1 ~ UnaryMinus: (Integer) -> Integer 3.74/1.84 < ~ Lt: (Integer, Integer) -> Boolean 3.74/1.84 || ~ Lor: (Boolean, Boolean) -> Boolean 3.74/1.84 - ~ Sub: (Integer, Integer) -> Integer 3.74/1.84 ~ ~ Bwnot: (Integer) -> Integer 3.74/1.84 * ~ Mul: (Integer, Integer) -> Integer 3.74/1.84 >>> 3.74/1.84 3.74/1.84 3.74/1.84 The following domains are used: 3.74/1.84 Boolean, Integer 3.74/1.84 3.74/1.84 The ITRS R consists of the following rules: 3.74/1.84 div(x, y) -> if(x >= y && y > 0, x, y) 3.74/1.84 if(TRUE, x, y) -> div(x - y, y) + 1 3.74/1.84 3.74/1.84 The integer pair graph contains the following rules and edges: 3.74/1.84 (0): DIV(x[0], y[0]) -> IF(x[0] >= y[0] && y[0] > 0, x[0], y[0]) 3.74/1.84 (1): IF(TRUE, x[1], y[1]) -> DIV(x[1] - y[1], y[1]) 3.74/1.84 3.74/1.84 (0) -> (1), if (x[0] >= y[0] && y[0] > 0 & x[0] ->^* x[1] & y[0] ->^* y[1]) 3.74/1.84 (1) -> (0), if (x[1] - y[1] ->^* x[0] & y[1] ->^* y[0]) 3.74/1.84 3.74/1.84 The set Q consists of the following terms: 3.74/1.84 div(x0, x1) 3.74/1.84 if(TRUE, x0, x1) 3.74/1.84 3.74/1.84 ---------------------------------------- 3.74/1.84 3.74/1.84 (3) UsableRulesProof (EQUIVALENT) 3.74/1.84 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. 3.74/1.84 ---------------------------------------- 3.74/1.84 3.74/1.84 (4) 3.74/1.84 Obligation: 3.74/1.84 IDP problem: 3.74/1.84 The following function symbols are pre-defined: 3.74/1.84 <<< 3.74/1.84 & ~ Bwand: (Integer, Integer) -> Integer 3.74/1.84 >= ~ Ge: (Integer, Integer) -> Boolean 3.74/1.84 | ~ Bwor: (Integer, Integer) -> Integer 3.74/1.84 / ~ Div: (Integer, Integer) -> Integer 3.74/1.84 != ~ Neq: (Integer, Integer) -> Boolean 3.74/1.84 && ~ Land: (Boolean, Boolean) -> Boolean 3.74/1.84 ! ~ Lnot: (Boolean) -> Boolean 3.74/1.84 = ~ Eq: (Integer, Integer) -> Boolean 3.74/1.84 <= ~ Le: (Integer, Integer) -> Boolean 3.74/1.84 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.74/1.84 % ~ Mod: (Integer, Integer) -> Integer 3.74/1.84 > ~ Gt: (Integer, Integer) -> Boolean 3.74/1.84 + ~ Add: (Integer, Integer) -> Integer 3.74/1.84 -1 ~ UnaryMinus: (Integer) -> Integer 3.74/1.84 < ~ Lt: (Integer, Integer) -> Boolean 3.74/1.84 || ~ Lor: (Boolean, Boolean) -> Boolean 3.74/1.84 - ~ Sub: (Integer, Integer) -> Integer 3.74/1.84 ~ ~ Bwnot: (Integer) -> Integer 3.74/1.84 * ~ Mul: (Integer, Integer) -> Integer 3.74/1.84 >>> 3.74/1.84 3.74/1.84 3.74/1.84 The following domains are used: 3.74/1.84 Boolean, Integer 3.74/1.84 3.74/1.84 R is empty. 3.74/1.84 3.74/1.84 The integer pair graph contains the following rules and edges: 3.74/1.84 (0): DIV(x[0], y[0]) -> IF(x[0] >= y[0] && y[0] > 0, x[0], y[0]) 3.74/1.84 (1): IF(TRUE, x[1], y[1]) -> DIV(x[1] - y[1], y[1]) 3.74/1.84 3.74/1.84 (0) -> (1), if (x[0] >= y[0] && y[0] > 0 & x[0] ->^* x[1] & y[0] ->^* y[1]) 3.74/1.84 (1) -> (0), if (x[1] - y[1] ->^* x[0] & y[1] ->^* y[0]) 3.74/1.84 3.74/1.84 The set Q consists of the following terms: 3.74/1.84 div(x0, x1) 3.74/1.84 if(TRUE, x0, x1) 3.74/1.84 3.74/1.84 ---------------------------------------- 3.74/1.84 3.74/1.84 (5) IDPNonInfProof (SOUND) 3.74/1.84 Used the following options for this NonInfProof: 3.74/1.84 3.74/1.84 IDPGPoloSolver: 3.74/1.84 Range: [(-1,2)] 3.74/1.84 IsNat: false 3.74/1.84 Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@74d0f7f 3.74/1.84 Constraint Generator: NonInfConstraintGenerator: 3.74/1.84 PathGenerator: MetricPathGenerator: 3.74/1.84 Max Left Steps: 1 3.74/1.84 Max Right Steps: 1 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 The constraints were generated the following way: 3.74/1.84 3.74/1.84 The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps: 3.74/1.84 3.74/1.84 Note that final constraints are written in bold face. 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 For Pair DIV(x, y) -> IF(&&(>=(x, y), >(y, 0)), x, y) the following chains were created: 3.74/1.84 *We consider the chain DIV(x[0], y[0]) -> IF(&&(>=(x[0], y[0]), >(y[0], 0)), x[0], y[0]), IF(TRUE, x[1], y[1]) -> DIV(-(x[1], y[1]), y[1]) which results in the following constraint: 3.74/1.84 3.74/1.84 (1) (&&(>=(x[0], y[0]), >(y[0], 0))=TRUE & x[0]=x[1] & y[0]=y[1] ==> DIV(x[0], y[0])_>=_NonInfC & DIV(x[0], y[0])_>=_IF(&&(>=(x[0], y[0]), >(y[0], 0)), x[0], y[0]) & (U^Increasing(IF(&&(>=(x[0], y[0]), >(y[0], 0)), x[0], y[0])), >=)) 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 We simplified constraint (1) using rules (IV), (IDP_BOOLEAN) which results in the following new constraint: 3.74/1.84 3.74/1.84 (2) (>=(x[0], y[0])=TRUE & >(y[0], 0)=TRUE ==> DIV(x[0], y[0])_>=_NonInfC & DIV(x[0], y[0])_>=_IF(&&(>=(x[0], y[0]), >(y[0], 0)), x[0], y[0]) & (U^Increasing(IF(&&(>=(x[0], y[0]), >(y[0], 0)), x[0], y[0])), >=)) 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 3.74/1.84 3.74/1.84 (3) (x[0] + [-1]y[0] >= 0 & y[0] + [-1] >= 0 ==> (U^Increasing(IF(&&(>=(x[0], y[0]), >(y[0], 0)), x[0], y[0])), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [(-1)bso_14] >= 0) 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 3.74/1.84 3.74/1.84 (4) (x[0] + [-1]y[0] >= 0 & y[0] + [-1] >= 0 ==> (U^Increasing(IF(&&(>=(x[0], y[0]), >(y[0], 0)), x[0], y[0])), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [(-1)bso_14] >= 0) 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 3.74/1.84 3.74/1.84 (5) (x[0] + [-1]y[0] >= 0 & y[0] + [-1] >= 0 ==> (U^Increasing(IF(&&(>=(x[0], y[0]), >(y[0], 0)), x[0], y[0])), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [(-1)bso_14] >= 0) 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 For Pair IF(TRUE, x, y) -> DIV(-(x, y), y) the following chains were created: 3.74/1.84 *We consider the chain DIV(x[0], y[0]) -> IF(&&(>=(x[0], y[0]), >(y[0], 0)), x[0], y[0]), IF(TRUE, x[1], y[1]) -> DIV(-(x[1], y[1]), y[1]), DIV(x[0], y[0]) -> IF(&&(>=(x[0], y[0]), >(y[0], 0)), x[0], y[0]) which results in the following constraint: 3.74/1.84 3.74/1.84 (1) (&&(>=(x[0], y[0]), >(y[0], 0))=TRUE & x[0]=x[1] & y[0]=y[1] & -(x[1], y[1])=x[0]1 & y[1]=y[0]1 ==> IF(TRUE, x[1], y[1])_>=_NonInfC & IF(TRUE, x[1], y[1])_>=_DIV(-(x[1], y[1]), y[1]) & (U^Increasing(DIV(-(x[1], y[1]), y[1])), >=)) 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 We simplified constraint (1) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint: 3.74/1.84 3.74/1.84 (2) (>=(x[0], y[0])=TRUE & >(y[0], 0)=TRUE ==> IF(TRUE, x[0], y[0])_>=_NonInfC & IF(TRUE, x[0], y[0])_>=_DIV(-(x[0], y[0]), y[0]) & (U^Increasing(DIV(-(x[1], y[1]), y[1])), >=)) 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 3.74/1.84 3.74/1.84 (3) (x[0] + [-1]y[0] >= 0 & y[0] + [-1] >= 0 ==> (U^Increasing(DIV(-(x[1], y[1]), y[1])), >=) & [(-1)bni_15 + (-1)Bound*bni_15] + [(-1)bni_15]y[0] + [bni_15]x[0] >= 0 & [(-1)bso_16] + y[0] >= 0) 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 3.74/1.84 3.74/1.84 (4) (x[0] + [-1]y[0] >= 0 & y[0] + [-1] >= 0 ==> (U^Increasing(DIV(-(x[1], y[1]), y[1])), >=) & [(-1)bni_15 + (-1)Bound*bni_15] + [(-1)bni_15]y[0] + [bni_15]x[0] >= 0 & [(-1)bso_16] + y[0] >= 0) 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 3.74/1.84 3.74/1.84 (5) (x[0] + [-1]y[0] >= 0 & y[0] + [-1] >= 0 ==> (U^Increasing(DIV(-(x[1], y[1]), y[1])), >=) & [(-1)bni_15 + (-1)Bound*bni_15] + [(-1)bni_15]y[0] + [bni_15]x[0] >= 0 & [(-1)bso_16] + y[0] >= 0) 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 To summarize, we get the following constraints P__>=_ for the following pairs. 3.74/1.84 3.74/1.84 *DIV(x, y) -> IF(&&(>=(x, y), >(y, 0)), x, y) 3.74/1.84 3.74/1.84 *(x[0] + [-1]y[0] >= 0 & y[0] + [-1] >= 0 ==> (U^Increasing(IF(&&(>=(x[0], y[0]), >(y[0], 0)), x[0], y[0])), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [(-1)bso_14] >= 0) 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 *IF(TRUE, x, y) -> DIV(-(x, y), y) 3.74/1.84 3.74/1.84 *(x[0] + [-1]y[0] >= 0 & y[0] + [-1] >= 0 ==> (U^Increasing(DIV(-(x[1], y[1]), y[1])), >=) & [(-1)bni_15 + (-1)Bound*bni_15] + [(-1)bni_15]y[0] + [bni_15]x[0] >= 0 & [(-1)bso_16] + y[0] >= 0) 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 3.74/1.84 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. 3.74/1.84 3.74/1.84 Using the following integer polynomial ordering the resulting constraints can be solved 3.74/1.84 3.74/1.84 Polynomial interpretation over integers[POLO]: 3.74/1.84 3.74/1.84 POL(TRUE) = 0 3.74/1.84 POL(FALSE) = [3] 3.74/1.84 POL(DIV(x_1, x_2)) = [-1] + [-1]x_2 + x_1 3.74/1.84 POL(IF(x_1, x_2, x_3)) = [-1] + [-1]x_3 + x_2 + [-1]x_1 3.74/1.84 POL(&&(x_1, x_2)) = 0 3.74/1.84 POL(>=(x_1, x_2)) = [-1] 3.74/1.84 POL(>(x_1, x_2)) = [-1] 3.74/1.84 POL(0) = 0 3.74/1.84 POL(-(x_1, x_2)) = x_1 + [-1]x_2 3.74/1.84 3.74/1.84 3.74/1.84 The following pairs are in P_>: 3.74/1.84 3.74/1.84 3.74/1.84 IF(TRUE, x[1], y[1]) -> DIV(-(x[1], y[1]), y[1]) 3.74/1.84 3.74/1.84 3.74/1.84 The following pairs are in P_bound: 3.74/1.84 3.74/1.84 3.74/1.84 DIV(x[0], y[0]) -> IF(&&(>=(x[0], y[0]), >(y[0], 0)), x[0], y[0]) 3.74/1.84 IF(TRUE, x[1], y[1]) -> DIV(-(x[1], y[1]), y[1]) 3.74/1.84 3.74/1.84 3.74/1.84 The following pairs are in P_>=: 3.74/1.84 3.74/1.84 3.74/1.84 DIV(x[0], y[0]) -> IF(&&(>=(x[0], y[0]), >(y[0], 0)), x[0], y[0]) 3.74/1.84 3.74/1.84 3.74/1.84 At least the following rules have been oriented under context sensitive arithmetic replacement: 3.74/1.84 3.74/1.84 TRUE^1 -> &&(TRUE, TRUE)^1 3.74/1.84 FALSE^1 -> &&(TRUE, FALSE)^1 3.74/1.84 FALSE^1 -> &&(FALSE, TRUE)^1 3.74/1.84 FALSE^1 -> &&(FALSE, FALSE)^1 3.74/1.84 3.74/1.84 ---------------------------------------- 3.74/1.84 3.74/1.84 (6) 3.74/1.84 Obligation: 3.74/1.84 IDP problem: 3.74/1.84 The following function symbols are pre-defined: 3.74/1.84 <<< 3.74/1.84 & ~ Bwand: (Integer, Integer) -> Integer 3.74/1.84 >= ~ Ge: (Integer, Integer) -> Boolean 3.74/1.84 | ~ Bwor: (Integer, Integer) -> Integer 3.74/1.84 / ~ Div: (Integer, Integer) -> Integer 3.74/1.84 != ~ Neq: (Integer, Integer) -> Boolean 3.74/1.84 && ~ Land: (Boolean, Boolean) -> Boolean 3.74/1.84 ! ~ Lnot: (Boolean) -> Boolean 3.74/1.84 = ~ Eq: (Integer, Integer) -> Boolean 3.74/1.84 <= ~ Le: (Integer, Integer) -> Boolean 3.74/1.84 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.74/1.84 % ~ Mod: (Integer, Integer) -> Integer 3.74/1.84 > ~ Gt: (Integer, Integer) -> Boolean 3.74/1.84 + ~ Add: (Integer, Integer) -> Integer 3.74/1.84 -1 ~ UnaryMinus: (Integer) -> Integer 3.74/1.84 < ~ Lt: (Integer, Integer) -> Boolean 3.74/1.84 || ~ Lor: (Boolean, Boolean) -> Boolean 3.74/1.84 - ~ Sub: (Integer, Integer) -> Integer 3.74/1.84 ~ ~ Bwnot: (Integer) -> Integer 3.74/1.84 * ~ Mul: (Integer, Integer) -> Integer 3.74/1.84 >>> 3.74/1.84 3.74/1.84 3.74/1.84 The following domains are used: 3.74/1.84 Boolean, Integer 3.74/1.84 3.74/1.84 R is empty. 3.74/1.84 3.74/1.84 The integer pair graph contains the following rules and edges: 3.74/1.84 (0): DIV(x[0], y[0]) -> IF(x[0] >= y[0] && y[0] > 0, x[0], y[0]) 3.74/1.84 3.74/1.84 3.74/1.84 The set Q consists of the following terms: 3.74/1.84 div(x0, x1) 3.74/1.84 if(TRUE, x0, x1) 3.74/1.84 3.74/1.84 ---------------------------------------- 3.74/1.84 3.74/1.84 (7) IDependencyGraphProof (EQUIVALENT) 3.74/1.84 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. 3.74/1.84 ---------------------------------------- 3.74/1.84 3.74/1.84 (8) 3.74/1.84 TRUE 3.74/1.87 EOF