/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. 0 : [] --> o 1 : [] --> o 2 : [] --> o f : [o * o] --> o g : [o * o] --> o i : [o] --> o f(0, X) => X f(X, 0) => X f(i(X), Y) => i(X) f(f(X, Y), Z) => f(X, f(Y, Z)) f(g(X, Y), Z) => g(f(X, Z), f(Y, Z)) f(1, g(X, Y)) => X f(2, g(X, Y)) => 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]): f(0, X) >? X f(X, 0) >? X f(i(X), Y) >? i(X) f(f(X, Y), Z) >? f(X, f(Y, Z)) f(g(X, Y), Z) >? g(f(X, Z), f(Y, Z)) f(1, g(X, Y)) >? X f(2, g(X, Y)) >? Y about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {f} and Mul = {0, 1, 2, g, i}, and the following precedence: i > 1 > 0 > f > g > 2 With these choices, we have: 1] f(0, X) >= X because [2], by (Star) 2] f*(0, X) >= X because [3], by (Select) 3] X >= X by (Meta) 4] f(X, 0) >= X because [5], by (Star) 5] f*(X, 0) >= X because [6], by (Select) 6] X >= X by (Meta) 7] f(i(X), Y) > i(X) because [8], by definition 8] f*(i(X), Y) >= i(X) because [9], by (Select) 9] i(X) >= i(X) because i in Mul and [10], by (Fun) 10] X >= X by (Meta) 11] f(f(X, Y), Z) >= f(X, f(Y, Z)) because [12], by (Star) 12] f*(f(X, Y), Z) >= f(X, f(Y, Z)) because [13], [15] and [17], by (Stat) 13] f(X, Y) > X because [14], by definition 14] f*(X, Y) >= X because [10], by (Select) 15] f*(f(X, Y), Z) >= X because [16], by (Select) 16] f(X, Y) >= X because [14], by (Star) 17] f*(f(X, Y), Z) >= f(Y, Z) because [18], [20] and [22], by (Stat) 18] f(X, Y) > Y because [19], by definition 19] f*(X, Y) >= Y because [3], by (Select) 20] f*(f(X, Y), Z) >= Y because [21], by (Select) 21] f(X, Y) >= Y because [19], by (Star) 22] f*(f(X, Y), Z) >= Z because [23], by (Select) 23] Z >= Z by (Meta) 24] f(g(X, Y), Z) >= g(f(X, Z), f(Y, Z)) because [25], by (Star) 25] f*(g(X, Y), Z) >= g(f(X, Z), f(Y, Z)) because f > g, [26] and [32], by (Copy) 26] f*(g(X, Y), Z) >= f(X, Z) because [27], [29] and [31], by (Stat) 27] g(X, Y) > X because [28], by definition 28] g*(X, Y) >= X because [10], by (Select) 29] f*(g(X, Y), Z) >= X because [30], by (Select) 30] g(X, Y) >= X because [28], by (Star) 31] f*(g(X, Y), Z) >= Z because [23], by (Select) 32] f*(g(X, Y), Z) >= f(Y, Z) because [33], [35] and [31], by (Stat) 33] g(X, Y) > Y because [34], by definition 34] g*(X, Y) >= Y because [3], by (Select) 35] f*(g(X, Y), Z) >= Y because [36], by (Select) 36] g(X, Y) >= Y because [34], by (Star) 37] f(1, g(X, Y)) > X because [38], by definition 38] f*(1, g(X, Y)) >= X because [30], by (Select) 39] f(2, g(X, Y)) >= Y because [40], by (Star) 40] f*(2, g(X, Y)) >= Y because [36], by (Select) We can thus remove the following rules: f(i(X), Y) => i(X) f(1, g(X, Y)) => 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]): f(0, X) >? X f(X, 0) >? X f(f(X, Y), Z) >? f(X, f(Y, Z)) f(g(X, Y), Z) >? g(f(X, Z), f(Y, Z)) f(2, g(X, Y)) >? Y about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {f} and Mul = {0, 2, g}, and the following precedence: 2 > f > g > 0 With these choices, we have: 1] f(0, X) >= X because [2], by (Star) 2] f*(0, X) >= X because [3], by (Select) 3] X >= X by (Meta) 4] f(X, 0) >= X because [5], by (Star) 5] f*(X, 0) >= X because [6], by (Select) 6] X >= X by (Meta) 7] f(f(X, Y), Z) >= f(X, f(Y, Z)) because [8], by (Star) 8] f*(f(X, Y), Z) >= f(X, f(Y, Z)) because [9], [11] and [13], by (Stat) 9] f(X, Y) > X because [10], by definition 10] f*(X, Y) >= X because [6], by (Select) 11] f*(f(X, Y), Z) >= X because [12], by (Select) 12] f(X, Y) >= X because [10], by (Star) 13] f*(f(X, Y), Z) >= f(Y, Z) because [14], [16] and [18], by (Stat) 14] f(X, Y) > Y because [15], by definition 15] f*(X, Y) >= Y because [3], by (Select) 16] f*(f(X, Y), Z) >= Y because [17], by (Select) 17] f(X, Y) >= Y because [15], by (Star) 18] f*(f(X, Y), Z) >= Z because [19], by (Select) 19] Z >= Z by (Meta) 20] f(g(X, Y), Z) > g(f(X, Z), f(Y, Z)) because [21], by definition 21] f*(g(X, Y), Z) >= g(f(X, Z), f(Y, Z)) because f > g, [22] and [28], by (Copy) 22] f*(g(X, Y), Z) >= f(X, Z) because [23], [25] and [27], by (Stat) 23] g(X, Y) > X because [24], by definition 24] g*(X, Y) >= X because [6], by (Select) 25] f*(g(X, Y), Z) >= X because [26], by (Select) 26] g(X, Y) >= X because [24], by (Star) 27] f*(g(X, Y), Z) >= Z because [19], by (Select) 28] f*(g(X, Y), Z) >= f(Y, Z) because [29], [31] and [27], by (Stat) 29] g(X, Y) > Y because [30], by definition 30] g*(X, Y) >= Y because [3], by (Select) 31] f*(g(X, Y), Z) >= Y because [32], by (Select) 32] g(X, Y) >= Y because [30], by (Star) 33] f(2, g(X, Y)) >= Y because [34], by (Star) 34] f*(2, g(X, Y)) >= Y because [32], by (Select) We can thus remove the following rules: f(g(X, Y), Z) => g(f(X, Z), f(Y, Z)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): f(0, X) >? X f(X, 0) >? X f(f(X, Y), Z) >? f(X, f(Y, Z)) f(2, g(X, Y)) >? Y We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 2 = 3 f = \y0y1.3 + y1 + 3y0 g = \y0y1.3 + y0 + y1 Using this interpretation, the requirements translate to: [[f(0, _x0)]] = 12 + x0 > x0 = [[_x0]] [[f(_x0, 0)]] = 6 + 3x0 > x0 = [[_x0]] [[f(f(_x0, _x1), _x2)]] = 12 + x2 + 3x1 + 9x0 > 6 + x2 + 3x0 + 3x1 = [[f(_x0, f(_x1, _x2))]] [[f(2, g(_x0, _x1))]] = 15 + x0 + x1 > x1 = [[_x1]] We can thus remove the following rules: f(0, X) => X f(X, 0) => X f(f(X, Y), Z) => f(X, f(Y, Z)) f(2, g(X, Y)) => 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.