/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. 0 : [] --> o activate : [o] --> o cons : [o * o] --> o from : [o] --> o n!6220!6220from : [o] --> o n!6220!6220s : [o] --> o s : [o] --> o sel : [o * o] --> o from(X) => cons(X, n!6220!6220from(n!6220!6220s(X))) sel(0, cons(X, Y)) => X sel(s(X), cons(Y, Z)) => sel(X, activate(Z)) from(X) => n!6220!6220from(X) s(X) => n!6220!6220s(X) activate(n!6220!6220from(X)) => from(activate(X)) activate(n!6220!6220s(X)) => s(activate(X)) activate(X) => 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]): from(X) >? cons(X, n!6220!6220from(n!6220!6220s(X))) sel(0, cons(X, Y)) >? X sel(s(X), cons(Y, Z)) >? sel(X, activate(Z)) from(X) >? n!6220!6220from(X) s(X) >? n!6220!6220s(X) activate(n!6220!6220from(X)) >? from(activate(X)) activate(n!6220!6220s(X)) >? s(activate(X)) activate(X) >? X about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {sel} and Mul = {0, activate, cons, from, n!6220!6220from, n!6220!6220s, s}, and the following precedence: 0 > sel > activate > from > cons > n!6220!6220from > s > n!6220!6220s With these choices, we have: 1] from(X) > cons(X, n!6220!6220from(n!6220!6220s(X))) because [2], by definition 2] from*(X) >= cons(X, n!6220!6220from(n!6220!6220s(X))) because from > cons, [3] and [5], by (Copy) 3] from*(X) >= X because [4], by (Select) 4] X >= X by (Meta) 5] from*(X) >= n!6220!6220from(n!6220!6220s(X)) because from > n!6220!6220from and [6], by (Copy) 6] from*(X) >= n!6220!6220s(X) because from > n!6220!6220s and [3], by (Copy) 7] sel(0, cons(X, Y)) > X because [8], by definition 8] sel*(0, cons(X, Y)) >= X because [9], by (Select) 9] cons(X, Y) >= X because [10], by (Star) 10] cons*(X, Y) >= X because [4], by (Select) 11] sel(s(X), cons(Y, Z)) >= sel(X, activate(Z)) because [12], by (Star) 12] sel*(s(X), cons(Y, Z)) >= sel(X, activate(Z)) because [13], [15] and [17], by (Stat) 13] s(X) > X because [14], by definition 14] s*(X) >= X because [4], by (Select) 15] sel*(s(X), cons(Y, Z)) >= X because [16], by (Select) 16] s(X) >= X because [14], by (Star) 17] sel*(s(X), cons(Y, Z)) >= activate(Z) because sel > activate and [18], by (Copy) 18] sel*(s(X), cons(Y, Z)) >= Z because [19], by (Select) 19] cons(Y, Z) >= Z because [20], by (Star) 20] cons*(Y, Z) >= Z because [21], by (Select) 21] Z >= Z by (Meta) 22] from(X) >= n!6220!6220from(X) because [23], by (Star) 23] from*(X) >= n!6220!6220from(X) because from > n!6220!6220from and [3], by (Copy) 24] s(X) > n!6220!6220s(X) because [25], by definition 25] s*(X) >= n!6220!6220s(X) because s > n!6220!6220s and [14], by (Copy) 26] activate(n!6220!6220from(X)) > from(activate(X)) because [27], by definition 27] activate*(n!6220!6220from(X)) >= from(activate(X)) because activate > from and [28], by (Copy) 28] activate*(n!6220!6220from(X)) >= activate(X) because activate in Mul and [29], by (Stat) 29] n!6220!6220from(X) > X because [30], by definition 30] n!6220!6220from*(X) >= X because [4], by (Select) 31] activate(n!6220!6220s(X)) > s(activate(X)) because [32], by definition 32] activate*(n!6220!6220s(X)) >= s(activate(X)) because activate > s and [33], by (Copy) 33] activate*(n!6220!6220s(X)) >= activate(X) because activate in Mul and [34], by (Stat) 34] n!6220!6220s(X) > X because [35], by definition 35] n!6220!6220s*(X) >= X because [4], by (Select) 36] activate(X) > X because [37], by definition 37] activate*(X) >= X because [4], by (Select) We can thus remove the following rules: from(X) => cons(X, n!6220!6220from(n!6220!6220s(X))) sel(0, cons(X, Y)) => X s(X) => n!6220!6220s(X) activate(n!6220!6220from(X)) => from(activate(X)) activate(n!6220!6220s(X)) => s(activate(X)) activate(X) => 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]): sel(s(X), cons(Y, Z)) >? sel(X, activate(Z)) from(X) >? n!6220!6220from(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: activate = \y0.y0 cons = \y0y1.3 + y0 + 3y1 from = \y0.3 + 3y0 n!6220!6220from = \y0.y0 s = \y0.3 + 3y0 sel = \y0y1.y0 + y1 Using this interpretation, the requirements translate to: [[sel(s(_x0), cons(_x1, _x2))]] = 6 + x1 + 3x0 + 3x2 > x0 + x2 = [[sel(_x0, activate(_x2))]] [[from(_x0)]] = 3 + 3x0 > x0 = [[n!6220!6220from(_x0)]] We can thus remove the following rules: sel(s(X), cons(Y, Z)) => sel(X, activate(Z)) from(X) => n!6220!6220from(X) 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.