/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 after : [o * o] --> o cons : [o * o] --> o from : [o] --> o n!6220!6220from : [o] --> o n!6220!6220s : [o] --> o s : [o] --> o from(X) => cons(X, n!6220!6220from(n!6220!6220s(X))) after(0, X) => X after(s(X), cons(Y, Z)) => after(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))) after(0, X) >? X after(s(X), cons(Y, Z)) >? after(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 = {after} and Mul = {0, activate, cons, from, n!6220!6220from, n!6220!6220s, s}, and the following precedence: 0 > after > 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] after(0, X) >= X because [8], by (Star) 8] after*(0, X) >= X because [9], by (Select) 9] X >= X by (Meta) 10] after(s(X), cons(Y, Z)) >= after(X, activate(Z)) because [11], by (Star) 11] after*(s(X), cons(Y, Z)) >= after(X, activate(Z)) because [12], [15] and [17], by (Stat) 12] s(X) > X because [13], by definition 13] s*(X) >= X because [14], by (Select) 14] X >= X by (Meta) 15] after*(s(X), cons(Y, Z)) >= X because [16], by (Select) 16] s(X) >= X because [13], by (Star) 17] after*(s(X), cons(Y, Z)) >= activate(Z) because after > activate and [18], by (Copy) 18] after*(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 [9], by (Select) 21] from(X) > n!6220!6220from(X) because [22], by definition 22] from*(X) >= n!6220!6220from(X) because from > n!6220!6220from and [3], by (Copy) 23] s(X) > n!6220!6220s(X) because [24], by definition 24] s*(X) >= n!6220!6220s(X) because s > n!6220!6220s and [25], by (Copy) 25] s*(X) >= X because [4], by (Select) 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 (Star) 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))) from(X) => n!6220!6220from(X) s(X) => n!6220!6220s(X) activate(n!6220!6220from(X)) => from(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]): after(0, X) >? X after(s(X), cons(Y, Z)) >? after(X, activate(Z)) activate(n!6220!6220s(X)) >? s(activate(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 activate = \y0.2 + 2y0 after = \y0y1.3 + y0 + y1 cons = \y0y1.3 + y0 + 3y1 n!6220!6220s = \y0.3 + 3y0 s = \y0.y0 Using this interpretation, the requirements translate to: [[after(0, _x0)]] = 6 + x0 > x0 = [[_x0]] [[after(s(_x0), cons(_x1, _x2))]] = 6 + x0 + x1 + 3x2 > 5 + x0 + 2x2 = [[after(_x0, activate(_x2))]] [[activate(n!6220!6220s(_x0))]] = 8 + 6x0 > 2 + 2x0 = [[s(activate(_x0))]] We can thus remove the following rules: after(0, X) => X after(s(X), cons(Y, Z)) => after(X, activate(Z)) activate(n!6220!6220s(X)) => s(activate(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.