/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. !minus : [o * o] --> o !plus : [o * o] --> o !times : [o * o] --> o 0 : [] --> o exp : [o * o] --> o s : [o] --> o exp(X, 0) => s(0) exp(X, s(Y)) => !times(X, exp(X, Y)) !times(0, X) => 0 !times(s(X), Y) => !plus(Y, !times(X, Y)) !minus(0, X) => 0 !minus(X, 0) => X !minus(s(X), s(Y)) => !minus(X, Y) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): exp(X, 0) >? s(0) exp(X, s(Y)) >? !times(X, exp(X, Y)) !times(0, X) >? 0 !times(s(X), Y) >? !plus(Y, !times(X, Y)) !minus(0, X) >? 0 !minus(X, 0) >? X !minus(s(X), s(Y)) >? !minus(X, Y) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {} and Mul = {!minus, !plus, !times, 0, exp, s}, and the following precedence: exp > !minus > s > !times > !plus > 0 With these choices, we have: 1] exp(X, 0) > s(0) because [2], by definition 2] exp*(X, 0) >= s(0) because exp > s and [3], by (Copy) 3] exp*(X, 0) >= 0 because exp > 0, by (Copy) 4] exp(X, s(Y)) >= !times(X, exp(X, Y)) because [5], by (Star) 5] exp*(X, s(Y)) >= !times(X, exp(X, Y)) because exp > !times, [6] and [8], by (Copy) 6] exp*(X, s(Y)) >= X because [7], by (Select) 7] X >= X by (Meta) 8] exp*(X, s(Y)) >= exp(X, Y) because exp in Mul, [9] and [10], by (Stat) 9] X >= X by (Meta) 10] s(Y) > Y because [11], by definition 11] s*(Y) >= Y because [12], by (Select) 12] Y >= Y by (Meta) 13] !times(0, X) > 0 because [14], by definition 14] !times*(0, X) >= 0 because !times > 0, by (Copy) 15] !times(s(X), Y) >= !plus(Y, !times(X, Y)) because [16], by (Star) 16] !times*(s(X), Y) >= !plus(Y, !times(X, Y)) because !times > !plus, [17] and [18], by (Copy) 17] !times*(s(X), Y) >= Y because [12], by (Select) 18] !times*(s(X), Y) >= !times(X, Y) because !times in Mul, [19] and [21], by (Stat) 19] s(X) > X because [20], by definition 20] s*(X) >= X because [9], by (Select) 21] Y >= Y by (Meta) 22] !minus(0, X) > 0 because [23], by definition 23] !minus*(0, X) >= 0 because [24], by (Select) 24] 0 >= 0 by (Fun) 25] !minus(X, 0) >= X because [26], by (Star) 26] !minus*(X, 0) >= X because [9], by (Select) 27] !minus(s(X), s(Y)) >= !minus(X, Y) because [28], by (Star) 28] !minus*(s(X), s(Y)) >= !minus(X, Y) because !minus in Mul, [29] and [10], by (Stat) 29] s(X) >= X because [20], by (Star) We can thus remove the following rules: exp(X, 0) => s(0) !times(0, X) => 0 !minus(0, X) => 0 We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): exp(X, s(Y)) >? !times(X, exp(X, Y)) !times(s(X), Y) >? !plus(Y, !times(X, Y)) !minus(X, 0) >? X !minus(s(X), s(Y)) >? !minus(X, Y) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {} and Mul = {!minus, !plus, !times, 0, exp, s}, and the following precedence: exp > 0 > s > !times > !plus > !minus With these choices, we have: 1] exp(X, s(Y)) > !times(X, exp(X, Y)) because [2], by definition 2] exp*(X, s(Y)) >= !times(X, exp(X, Y)) because exp > !times, [3] and [5], by (Copy) 3] exp*(X, s(Y)) >= X because [4], by (Select) 4] X >= X by (Meta) 5] exp*(X, s(Y)) >= exp(X, Y) because exp in Mul, [6] and [7], by (Stat) 6] X >= X by (Meta) 7] s(Y) > Y because [8], by definition 8] s*(Y) >= Y because [9], by (Select) 9] Y >= Y by (Meta) 10] !times(s(X), Y) >= !plus(Y, !times(X, Y)) because [11], by (Star) 11] !times*(s(X), Y) >= !plus(Y, !times(X, Y)) because !times > !plus, [12] and [13], by (Copy) 12] !times*(s(X), Y) >= Y because [9], by (Select) 13] !times*(s(X), Y) >= !times(X, Y) because !times in Mul, [14] and [16], by (Stat) 14] s(X) > X because [15], by definition 15] s*(X) >= X because [6], by (Select) 16] Y >= Y by (Meta) 17] !minus(X, 0) > X because [18], by definition 18] !minus*(X, 0) >= X because [6], by (Select) 19] !minus(s(X), s(Y)) >= !minus(X, Y) because [20], by (Star) 20] !minus*(s(X), s(Y)) >= !minus(X, Y) because !minus in Mul, [14] and [21], by (Stat) 21] s(Y) >= Y because [8], by (Star) We can thus remove the following rules: exp(X, s(Y)) => !times(X, exp(X, Y)) !minus(X, 0) => X We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): !times(s(X), Y) >? !plus(Y, !times(X, Y)) !minus(s(X), s(Y)) >? !minus(X, Y) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {} and Mul = {!minus, !plus, !times, s}, and the following precedence: s > !times > !plus > !minus With these choices, we have: 1] !times(s(X), Y) > !plus(Y, !times(X, Y)) because [2], by definition 2] !times*(s(X), Y) >= !plus(Y, !times(X, Y)) because !times > !plus, [3] and [5], by (Copy) 3] !times*(s(X), Y) >= Y because [4], by (Select) 4] Y >= Y by (Meta) 5] !times*(s(X), Y) >= !times(X, Y) because !times in Mul, [6] and [9], by (Stat) 6] s(X) > X because [7], by definition 7] s*(X) >= X because [8], by (Select) 8] X >= X by (Meta) 9] Y >= Y by (Meta) 10] !minus(s(X), s(Y)) >= !minus(X, Y) because !minus in Mul, [11] and [12], by (Fun) 11] s(X) >= X because [7], by (Star) 12] s(Y) >= Y because [13], by (Star) 13] s*(Y) >= Y because [9], by (Select) We can thus remove the following rules: !times(s(X), Y) => !plus(Y, !times(X, Y)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): !minus(s(X), s(Y)) >? !minus(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !minus = \y0y1.y0 + y1 s = \y0.3 + 3y0 Using this interpretation, the requirements translate to: [[!minus(s(_x0), s(_x1))]] = 6 + 3x0 + 3x1 > x0 + x1 = [[!minus(_x0, _x1)]] We can thus remove the following rules: !minus(s(X), s(Y)) => !minus(X, Y) All rules were succesfully removed. Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012.