/export/starexec/sandbox2/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. Alphabet: 0 : [] --> nat add : [nat] --> nat -> nat eq : [nat] --> nat -> bool err : [] --> nat false : [] --> bool id : [] --> nat -> nat nul : [] --> nat -> bool pred : [nat] --> nat s : [nat] --> nat true : [] --> bool Rules: nul 0 => true nul s(x) => false nul err => false pred(0) => err pred(s(x)) => x id x => x eq(0) => nul eq(s(x)) => /\y.eq(x) pred(y) add(0) => id add(s(x)) => /\y.add(x) s(y) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 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]): nul 0 >? true nul s(X) >? false nul err >? false pred(0) >? err pred(s(X)) >? X id X >? X eq(0) >? nul eq(s(X)) >? /\x.eq(X) pred(x) add(0) >? id add(s(X)) >? /\x.add(X) s(x) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[false]] = _|_ [[id]] = _|_ [[nul]] = _|_ [[true]] = _|_ We choose Lex = {} and Mul = {0, @_{o -> o}, add, eq, err, pred, s}, and the following precedence: add > 0 > s > eq > pred > @_{o -> o} > err Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: @_{o -> o}(_|_, 0) >= _|_ @_{o -> o}(_|_, s(X)) >= _|_ @_{o -> o}(_|_, err) >= _|_ pred(0) >= err pred(s(X)) > X @_{o -> o}(_|_, X) >= X eq(0) >= _|_ eq(s(X)) >= /\x.@_{o -> o}(eq(X), pred(x)) add(0) >= _|_ add(s(X)) >= /\x.@_{o -> o}(add(X), s(x)) With these choices, we have: 1] @_{o -> o}(_|_, 0) >= _|_ by (Bot) 2] @_{o -> o}(_|_, s(X)) >= _|_ by (Bot) 3] @_{o -> o}(_|_, err) >= _|_ by (Bot) 4] pred(0) >= err because [5], by (Star) 5] pred*(0) >= err because pred > err, by (Copy) 6] pred(s(X)) > X because [7], by definition 7] pred*(s(X)) >= X because [8], by (Select) 8] s(X) >= X because [9], by (Star) 9] s*(X) >= X because [10], by (Select) 10] X >= X by (Meta) 11] @_{o -> o}(_|_, X) >= X because [12], by (Star) 12] @_{o -> o}*(_|_, X) >= X because [13], by (Select) 13] X >= X by (Meta) 14] eq(0) >= _|_ by (Bot) 15] eq(s(X)) >= /\x.@_{o -> o}(eq(X), pred(x)) because [16], by (Star) 16] eq*(s(X)) >= /\y.@_{o -> o}(eq(X), pred(y)) because [17], by (F-Abs) 17] eq*(s(X), x) >= @_{o -> o}(eq(X), pred(x)) because eq > @_{o -> o}, [18] and [22], by (Copy) 18] eq*(s(X), x) >= eq(X) because eq in Mul and [19], by (Stat) 19] s(X) > X because [20], by definition 20] s*(X) >= X because [21], by (Select) 21] X >= X by (Meta) 22] eq*(s(X), x) >= pred(x) because eq > pred and [23], by (Copy) 23] eq*(s(X), x) >= x because [24], by (Select) 24] x >= x by (Var) 25] add(0) >= _|_ by (Bot) 26] add(s(X)) >= /\x.@_{o -> o}(add(X), s(x)) because [27], by (Star) 27] add*(s(X)) >= /\y.@_{o -> o}(add(X), s(y)) because [28], by (F-Abs) 28] add*(s(X), x) >= @_{o -> o}(add(X), s(x)) because add > @_{o -> o}, [29] and [33], by (Copy) 29] add*(s(X), x) >= add(X) because add in Mul and [30], by (Stat) 30] s(X) > X because [31], by definition 31] s*(X) >= X because [32], by (Select) 32] X >= X by (Meta) 33] add*(s(X), x) >= s(x) because add > s and [34], by (Copy) 34] add*(s(X), x) >= x because [35], by (Select) 35] x >= x by (Var) We can thus remove the following rules: pred(s(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]): nul 0 >? true nul s(X) >? false nul err >? false pred(0) >? err id X >? X eq(0) >? nul eq(s(X)) >? /\x.eq(X) pred(x) add(0) >? id add(s(X)) >? /\x.add(X) s(x) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[err]] = _|_ [[false]] = _|_ [[id]] = _|_ [[nul]] = _|_ [[true]] = _|_ We choose Lex = {} and Mul = {0, @_{o -> o}, add, eq, pred, s}, and the following precedence: 0 > eq > pred > add > @_{o -> o} > s Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: @_{o -> o}(_|_, 0) >= _|_ @_{o -> o}(_|_, s(X)) >= _|_ @_{o -> o}(_|_, _|_) >= _|_ pred(0) >= _|_ @_{o -> o}(_|_, X) >= X eq(0) >= _|_ eq(s(X)) > /\x.@_{o -> o}(eq(X), pred(x)) add(0) >= _|_ add(s(X)) > /\x.@_{o -> o}(add(X), s(x)) With these choices, we have: 1] @_{o -> o}(_|_, 0) >= _|_ by (Bot) 2] @_{o -> o}(_|_, s(X)) >= _|_ by (Bot) 3] @_{o -> o}(_|_, _|_) >= _|_ by (Bot) 4] pred(0) >= _|_ by (Bot) 5] @_{o -> o}(_|_, X) >= X because [6], by (Star) 6] @_{o -> o}*(_|_, X) >= X because [7], by (Select) 7] X >= X by (Meta) 8] eq(0) >= _|_ by (Bot) 9] eq(s(X)) > /\x.@_{o -> o}(eq(X), pred(x)) because [10], by definition 10] eq*(s(X)) >= /\y.@_{o -> o}(eq(X), pred(y)) because [11], by (F-Abs) 11] eq*(s(X), x) >= @_{o -> o}(eq(X), pred(x)) because eq > @_{o -> o}, [12] and [16], by (Copy) 12] eq*(s(X), x) >= eq(X) because eq in Mul and [13], by (Stat) 13] s(X) > X because [14], by definition 14] s*(X) >= X because [15], by (Select) 15] X >= X by (Meta) 16] eq*(s(X), x) >= pred(x) because eq > pred and [17], by (Copy) 17] eq*(s(X), x) >= x because [18], by (Select) 18] x >= x by (Var) 19] add(0) >= _|_ by (Bot) 20] add(s(X)) > /\x.@_{o -> o}(add(X), s(x)) because [21], by definition 21] add*(s(X)) >= /\y.@_{o -> o}(add(X), s(y)) because [22], by (F-Abs) 22] add*(s(X), x) >= @_{o -> o}(add(X), s(x)) because add > @_{o -> o}, [23] and [27], by (Copy) 23] add*(s(X), x) >= add(X) because add in Mul and [24], by (Stat) 24] s(X) > X because [25], by definition 25] s*(X) >= X because [26], by (Select) 26] X >= X by (Meta) 27] add*(s(X), x) >= s(x) because add > s and [28], by (Copy) 28] add*(s(X), x) >= x because [29], by (Select) 29] x >= x by (Var) We can thus remove the following rules: eq(s(X)) => /\x.eq(X) pred(x) add(s(X)) => /\x.add(X) s(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]): nul 0 >? true nul s(X) >? false nul err >? false pred(0) >? err id X >? X eq(0) >? nul add(0) >? id We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 add = \y0y1.3 + 3y0 + 3y1 eq = \y0y1.3 + 3y0 + 3y1 err = 0 false = 0 id = \y0.3y0 nul = \y0.3y0 pred = \y0.3 + 3y0 s = \y0.3 + y0 true = 0 Using this interpretation, the requirements translate to: [[nul 0]] = 12 > 0 = [[true]] [[nul s(_x0)]] = 12 + 4x0 > 0 = [[false]] [[nul err]] = 0 >= 0 = [[false]] [[pred(0)]] = 12 > 0 = [[err]] [[id _x0]] = 4x0 >= x0 = [[_x0]] [[eq(0)]] = \y0.12 + 3y0 > \y0.3y0 = [[nul]] [[add(0)]] = \y0.12 + 3y0 > \y0.3y0 = [[id]] We can thus remove the following rules: nul 0 => true nul s(X) => false pred(0) => err eq(0) => nul add(0) => id We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): nul(err) >? false id(X) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: err = 3 false = 0 id = \y0.3 + y0 nul = \y0.3 + 3y0 Using this interpretation, the requirements translate to: [[nul(err)]] = 12 > 0 = [[false]] [[id(_x0)]] = 3 + x0 > x0 = [[_x0]] We can thus remove the following rules: nul(err) => false id(X) => 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.