/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 a!6220!6220add : [o * o] --> o a!6220!6220dbl : [o] --> o a!6220!6220first : [o * o] --> o a!6220!6220sqr : [o] --> o a!6220!6220terms : [o] --> o add : [o * o] --> o cons : [o * o] --> o dbl : [o] --> o first : [o * o] --> o mark : [o] --> o nil : [] --> o recip : [o] --> o s : [o] --> o sqr : [o] --> o terms : [o] --> o a!6220!6220terms(X) => cons(recip(a!6220!6220sqr(mark(X))), terms(s(X))) a!6220!6220sqr(0) => 0 a!6220!6220sqr(s(X)) => s(add(sqr(X), dbl(X))) a!6220!6220dbl(0) => 0 a!6220!6220dbl(s(X)) => s(s(dbl(X))) a!6220!6220add(0, X) => mark(X) a!6220!6220add(s(X), Y) => s(add(X, Y)) a!6220!6220first(0, X) => nil a!6220!6220first(s(X), cons(Y, Z)) => cons(mark(Y), first(X, Z)) mark(terms(X)) => a!6220!6220terms(mark(X)) mark(sqr(X)) => a!6220!6220sqr(mark(X)) mark(add(X, Y)) => a!6220!6220add(mark(X), mark(Y)) mark(dbl(X)) => a!6220!6220dbl(mark(X)) mark(first(X, Y)) => a!6220!6220first(mark(X), mark(Y)) mark(cons(X, Y)) => cons(mark(X), Y) mark(recip(X)) => recip(mark(X)) mark(s(X)) => s(X) mark(0) => 0 mark(nil) => nil a!6220!6220terms(X) => terms(X) a!6220!6220sqr(X) => sqr(X) a!6220!6220add(X, Y) => add(X, Y) a!6220!6220dbl(X) => dbl(X) a!6220!6220first(X, Y) => first(X, Y) We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]). We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] a!6220!6220terms#(X) =#> a!6220!6220sqr#(mark(X)) 1] a!6220!6220terms#(X) =#> mark#(X) 2] a!6220!6220add#(0, X) =#> mark#(X) 3] a!6220!6220first#(s(X), cons(Y, Z)) =#> mark#(Y) 4] mark#(terms(X)) =#> a!6220!6220terms#(mark(X)) 5] mark#(terms(X)) =#> mark#(X) 6] mark#(sqr(X)) =#> a!6220!6220sqr#(mark(X)) 7] mark#(sqr(X)) =#> mark#(X) 8] mark#(add(X, Y)) =#> a!6220!6220add#(mark(X), mark(Y)) 9] mark#(add(X, Y)) =#> mark#(X) 10] mark#(add(X, Y)) =#> mark#(Y) 11] mark#(dbl(X)) =#> a!6220!6220dbl#(mark(X)) 12] mark#(dbl(X)) =#> mark#(X) 13] mark#(first(X, Y)) =#> a!6220!6220first#(mark(X), mark(Y)) 14] mark#(first(X, Y)) =#> mark#(X) 15] mark#(first(X, Y)) =#> mark#(Y) 16] mark#(cons(X, Y)) =#> mark#(X) 17] mark#(recip(X)) =#> mark#(X) Rules R_0: a!6220!6220terms(X) => cons(recip(a!6220!6220sqr(mark(X))), terms(s(X))) a!6220!6220sqr(0) => 0 a!6220!6220sqr(s(X)) => s(add(sqr(X), dbl(X))) a!6220!6220dbl(0) => 0 a!6220!6220dbl(s(X)) => s(s(dbl(X))) a!6220!6220add(0, X) => mark(X) a!6220!6220add(s(X), Y) => s(add(X, Y)) a!6220!6220first(0, X) => nil a!6220!6220first(s(X), cons(Y, Z)) => cons(mark(Y), first(X, Z)) mark(terms(X)) => a!6220!6220terms(mark(X)) mark(sqr(X)) => a!6220!6220sqr(mark(X)) mark(add(X, Y)) => a!6220!6220add(mark(X), mark(Y)) mark(dbl(X)) => a!6220!6220dbl(mark(X)) mark(first(X, Y)) => a!6220!6220first(mark(X), mark(Y)) mark(cons(X, Y)) => cons(mark(X), Y) mark(recip(X)) => recip(mark(X)) mark(s(X)) => s(X) mark(0) => 0 mark(nil) => nil a!6220!6220terms(X) => terms(X) a!6220!6220sqr(X) => sqr(X) a!6220!6220add(X, Y) => add(X, Y) a!6220!6220dbl(X) => dbl(X) a!6220!6220first(X, Y) => first(X, Y) Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_0, R_0, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : * 1 : 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 * 2 : 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 * 3 : 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 * 4 : 0, 1 * 5 : 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 * 6 : * 7 : 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 * 8 : 2 * 9 : 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 * 10 : 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 * 11 : * 12 : 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 * 13 : 3 * 14 : 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 * 15 : 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 * 16 : 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 * 17 : 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 This graph has the following strongly connected components: P_1: a!6220!6220terms#(X) =#> mark#(X) a!6220!6220add#(0, X) =#> mark#(X) a!6220!6220first#(s(X), cons(Y, Z)) =#> mark#(Y) mark#(terms(X)) =#> a!6220!6220terms#(mark(X)) mark#(terms(X)) =#> mark#(X) mark#(sqr(X)) =#> mark#(X) mark#(add(X, Y)) =#> a!6220!6220add#(mark(X), mark(Y)) mark#(add(X, Y)) =#> mark#(X) mark#(add(X, Y)) =#> mark#(Y) mark#(dbl(X)) =#> mark#(X) mark#(first(X, Y)) =#> a!6220!6220first#(mark(X), mark(Y)) mark#(first(X, Y)) =#> mark#(X) mark#(first(X, Y)) =#> mark#(Y) mark#(cons(X, Y)) =#> mark#(X) mark#(recip(X)) =#> mark#(X) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f). Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_0, minimal, formative). The formative rules of (P_1, R_0) are R_1 ::= a!6220!6220terms(X) => cons(recip(a!6220!6220sqr(mark(X))), terms(s(X))) a!6220!6220sqr(0) => 0 a!6220!6220sqr(s(X)) => s(add(sqr(X), dbl(X))) a!6220!6220dbl(0) => 0 a!6220!6220dbl(s(X)) => s(s(dbl(X))) a!6220!6220add(0, X) => mark(X) a!6220!6220add(s(X), Y) => s(add(X, Y)) a!6220!6220first(s(X), cons(Y, Z)) => cons(mark(Y), first(X, Z)) mark(terms(X)) => a!6220!6220terms(mark(X)) mark(sqr(X)) => a!6220!6220sqr(mark(X)) mark(add(X, Y)) => a!6220!6220add(mark(X), mark(Y)) mark(dbl(X)) => a!6220!6220dbl(mark(X)) mark(first(X, Y)) => a!6220!6220first(mark(X), mark(Y)) mark(cons(X, Y)) => cons(mark(X), Y) mark(recip(X)) => recip(mark(X)) mark(s(X)) => s(X) mark(0) => 0 a!6220!6220terms(X) => terms(X) a!6220!6220sqr(X) => sqr(X) a!6220!6220add(X, Y) => add(X, Y) a!6220!6220dbl(X) => dbl(X) a!6220!6220first(X, Y) => first(X, Y) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_1, R_1, minimal, formative). Thus, the original system is terminating if (P_1, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_1, minimal, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: a!6220!6220terms#(X) >? mark#(X) a!6220!6220add#(0, X) >? mark#(X) a!6220!6220first#(s(X), cons(Y, Z)) >? mark#(Y) mark#(terms(X)) >? a!6220!6220terms#(mark(X)) mark#(terms(X)) >? mark#(X) mark#(sqr(X)) >? mark#(X) mark#(add(X, Y)) >? a!6220!6220add#(mark(X), mark(Y)) mark#(add(X, Y)) >? mark#(X) mark#(add(X, Y)) >? mark#(Y) mark#(dbl(X)) >? mark#(X) mark#(first(X, Y)) >? a!6220!6220first#(mark(X), mark(Y)) mark#(first(X, Y)) >? mark#(X) mark#(first(X, Y)) >? mark#(Y) mark#(cons(X, Y)) >? mark#(X) mark#(recip(X)) >? mark#(X) a!6220!6220terms(X) >= cons(recip(a!6220!6220sqr(mark(X))), terms(s(X))) a!6220!6220sqr(0) >= 0 a!6220!6220sqr(s(X)) >= s(add(sqr(X), dbl(X))) a!6220!6220dbl(0) >= 0 a!6220!6220dbl(s(X)) >= s(s(dbl(X))) a!6220!6220add(0, X) >= mark(X) a!6220!6220add(s(X), Y) >= s(add(X, Y)) a!6220!6220first(s(X), cons(Y, Z)) >= cons(mark(Y), first(X, Z)) mark(terms(X)) >= a!6220!6220terms(mark(X)) mark(sqr(X)) >= a!6220!6220sqr(mark(X)) mark(add(X, Y)) >= a!6220!6220add(mark(X), mark(Y)) mark(dbl(X)) >= a!6220!6220dbl(mark(X)) mark(first(X, Y)) >= a!6220!6220first(mark(X), mark(Y)) mark(cons(X, Y)) >= cons(mark(X), Y) mark(recip(X)) >= recip(mark(X)) mark(s(X)) >= s(X) mark(0) >= 0 a!6220!6220terms(X) >= terms(X) a!6220!6220sqr(X) >= sqr(X) a!6220!6220add(X, Y) >= add(X, Y) a!6220!6220dbl(X) >= dbl(X) a!6220!6220first(X, Y) >= first(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 a!6220!6220add = \y0y1.y0 + 2y1 a!6220!6220add# = \y0y1.y1 a!6220!6220dbl = \y0.y0 a!6220!6220first = \y0y1.y0 + 2y1 a!6220!6220first# = \y0y1.y1 a!6220!6220sqr = \y0.y0 a!6220!6220terms = \y0.2 + 2y0 a!6220!6220terms# = \y0.y0 add = \y0y1.y0 + 2y1 cons = \y0y1.y0 dbl = \y0.y0 first = \y0y1.y0 + 2y1 mark = \y0.2y0 mark# = \y0.y0 recip = \y0.y0 s = \y0.0 sqr = \y0.y0 terms = \y0.2 + 2y0 Using this interpretation, the requirements translate to: [[a!6220!6220terms#(_x0)]] = x0 >= x0 = [[mark#(_x0)]] [[a!6220!6220add#(0, _x0)]] = x0 >= x0 = [[mark#(_x0)]] [[a!6220!6220first#(s(_x0), cons(_x1, _x2))]] = x1 >= x1 = [[mark#(_x1)]] [[mark#(terms(_x0))]] = 2 + 2x0 > 2x0 = [[a!6220!6220terms#(mark(_x0))]] [[mark#(terms(_x0))]] = 2 + 2x0 > x0 = [[mark#(_x0)]] [[mark#(sqr(_x0))]] = x0 >= x0 = [[mark#(_x0)]] [[mark#(add(_x0, _x1))]] = x0 + 2x1 >= 2x1 = [[a!6220!6220add#(mark(_x0), mark(_x1))]] [[mark#(add(_x0, _x1))]] = x0 + 2x1 >= x0 = [[mark#(_x0)]] [[mark#(add(_x0, _x1))]] = x0 + 2x1 >= x1 = [[mark#(_x1)]] [[mark#(dbl(_x0))]] = x0 >= x0 = [[mark#(_x0)]] [[mark#(first(_x0, _x1))]] = x0 + 2x1 >= 2x1 = [[a!6220!6220first#(mark(_x0), mark(_x1))]] [[mark#(first(_x0, _x1))]] = x0 + 2x1 >= x0 = [[mark#(_x0)]] [[mark#(first(_x0, _x1))]] = x0 + 2x1 >= x1 = [[mark#(_x1)]] [[mark#(cons(_x0, _x1))]] = x0 >= x0 = [[mark#(_x0)]] [[mark#(recip(_x0))]] = x0 >= x0 = [[mark#(_x0)]] [[a!6220!6220terms(_x0)]] = 2 + 2x0 >= 2x0 = [[cons(recip(a!6220!6220sqr(mark(_x0))), terms(s(_x0)))]] [[a!6220!6220sqr(0)]] = 0 >= 0 = [[0]] [[a!6220!6220sqr(s(_x0))]] = 0 >= 0 = [[s(add(sqr(_x0), dbl(_x0)))]] [[a!6220!6220dbl(0)]] = 0 >= 0 = [[0]] [[a!6220!6220dbl(s(_x0))]] = 0 >= 0 = [[s(s(dbl(_x0)))]] [[a!6220!6220add(0, _x0)]] = 2x0 >= 2x0 = [[mark(_x0)]] [[a!6220!6220add(s(_x0), _x1)]] = 2x1 >= 0 = [[s(add(_x0, _x1))]] [[a!6220!6220first(s(_x0), cons(_x1, _x2))]] = 2x1 >= 2x1 = [[cons(mark(_x1), first(_x0, _x2))]] [[mark(terms(_x0))]] = 4 + 4x0 >= 2 + 4x0 = [[a!6220!6220terms(mark(_x0))]] [[mark(sqr(_x0))]] = 2x0 >= 2x0 = [[a!6220!6220sqr(mark(_x0))]] [[mark(add(_x0, _x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[a!6220!6220add(mark(_x0), mark(_x1))]] [[mark(dbl(_x0))]] = 2x0 >= 2x0 = [[a!6220!6220dbl(mark(_x0))]] [[mark(first(_x0, _x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[a!6220!6220first(mark(_x0), mark(_x1))]] [[mark(cons(_x0, _x1))]] = 2x0 >= 2x0 = [[cons(mark(_x0), _x1)]] [[mark(recip(_x0))]] = 2x0 >= 2x0 = [[recip(mark(_x0))]] [[mark(s(_x0))]] = 0 >= 0 = [[s(_x0)]] [[mark(0)]] = 0 >= 0 = [[0]] [[a!6220!6220terms(_x0)]] = 2 + 2x0 >= 2 + 2x0 = [[terms(_x0)]] [[a!6220!6220sqr(_x0)]] = x0 >= x0 = [[sqr(_x0)]] [[a!6220!6220add(_x0, _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[add(_x0, _x1)]] [[a!6220!6220dbl(_x0)]] = x0 >= x0 = [[dbl(_x0)]] [[a!6220!6220first(_x0, _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[first(_x0, _x1)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_1, minimal, formative) by (P_2, R_1, minimal, formative), where P_2 consists of: a!6220!6220terms#(X) =#> mark#(X) a!6220!6220add#(0, X) =#> mark#(X) a!6220!6220first#(s(X), cons(Y, Z)) =#> mark#(Y) mark#(sqr(X)) =#> mark#(X) mark#(add(X, Y)) =#> a!6220!6220add#(mark(X), mark(Y)) mark#(add(X, Y)) =#> mark#(X) mark#(add(X, Y)) =#> mark#(Y) mark#(dbl(X)) =#> mark#(X) mark#(first(X, Y)) =#> a!6220!6220first#(mark(X), mark(Y)) mark#(first(X, Y)) =#> mark#(X) mark#(first(X, Y)) =#> mark#(Y) mark#(cons(X, Y)) =#> mark#(X) mark#(recip(X)) =#> mark#(X) Thus, the original system is terminating if (P_2, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_1, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 * 1 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 * 2 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 * 3 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 * 4 : 1 * 5 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 * 6 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 * 7 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 * 8 : 2 * 9 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 * 10 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 * 11 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 * 12 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 This graph has the following strongly connected components: P_3: a!6220!6220add#(0, X) =#> mark#(X) a!6220!6220first#(s(X), cons(Y, Z)) =#> mark#(Y) mark#(sqr(X)) =#> mark#(X) mark#(add(X, Y)) =#> a!6220!6220add#(mark(X), mark(Y)) mark#(add(X, Y)) =#> mark#(X) mark#(add(X, Y)) =#> mark#(Y) mark#(dbl(X)) =#> mark#(X) mark#(first(X, Y)) =#> a!6220!6220first#(mark(X), mark(Y)) mark#(first(X, Y)) =#> mark#(X) mark#(first(X, Y)) =#> mark#(Y) mark#(cons(X, Y)) =#> mark#(X) mark#(recip(X)) =#> mark#(X) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_2, R_1, m, f) by (P_3, R_1, m, f). Thus, the original system is terminating if (P_3, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_3, R_1, minimal, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: a!6220!6220add#(0, X) >? mark#(X) a!6220!6220first#(s(X), cons(Y, Z)) >? mark#(Y) mark#(sqr(X)) >? mark#(X) mark#(add(X, Y)) >? a!6220!6220add#(mark(X), mark(Y)) mark#(add(X, Y)) >? mark#(X) mark#(add(X, Y)) >? mark#(Y) mark#(dbl(X)) >? mark#(X) mark#(first(X, Y)) >? a!6220!6220first#(mark(X), mark(Y)) mark#(first(X, Y)) >? mark#(X) mark#(first(X, Y)) >? mark#(Y) mark#(cons(X, Y)) >? mark#(X) mark#(recip(X)) >? mark#(X) a!6220!6220terms(X) >= cons(recip(a!6220!6220sqr(mark(X))), terms(s(X))) a!6220!6220sqr(0) >= 0 a!6220!6220sqr(s(X)) >= s(add(sqr(X), dbl(X))) a!6220!6220dbl(0) >= 0 a!6220!6220dbl(s(X)) >= s(s(dbl(X))) a!6220!6220add(0, X) >= mark(X) a!6220!6220add(s(X), Y) >= s(add(X, Y)) a!6220!6220first(s(X), cons(Y, Z)) >= cons(mark(Y), first(X, Z)) mark(terms(X)) >= a!6220!6220terms(mark(X)) mark(sqr(X)) >= a!6220!6220sqr(mark(X)) mark(add(X, Y)) >= a!6220!6220add(mark(X), mark(Y)) mark(dbl(X)) >= a!6220!6220dbl(mark(X)) mark(first(X, Y)) >= a!6220!6220first(mark(X), mark(Y)) mark(cons(X, Y)) >= cons(mark(X), Y) mark(recip(X)) >= recip(mark(X)) mark(s(X)) >= s(X) mark(0) >= 0 a!6220!6220terms(X) >= terms(X) a!6220!6220sqr(X) >= sqr(X) a!6220!6220add(X, Y) >= add(X, Y) a!6220!6220dbl(X) >= dbl(X) a!6220!6220first(X, Y) >= first(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 a!6220!6220add = \y0y1.y0 + 2y1 a!6220!6220add# = \y0y1.y1 a!6220!6220dbl = \y0.y0 a!6220!6220first = \y0y1.y0 + 2y1 a!6220!6220first# = \y0y1.y1 a!6220!6220sqr = \y0.y0 a!6220!6220terms = \y0.2 + 2y0 add = \y0y1.y0 + 2y1 cons = \y0y1.1 + y0 dbl = \y0.y0 first = \y0y1.y0 + 2y1 mark = \y0.2y0 mark# = \y0.y0 recip = \y0.y0 s = \y0.0 sqr = \y0.y0 terms = \y0.2 + 2y0 Using this interpretation, the requirements translate to: [[a!6220!6220add#(0, _x0)]] = x0 >= x0 = [[mark#(_x0)]] [[a!6220!6220first#(s(_x0), cons(_x1, _x2))]] = 1 + x1 > x1 = [[mark#(_x1)]] [[mark#(sqr(_x0))]] = x0 >= x0 = [[mark#(_x0)]] [[mark#(add(_x0, _x1))]] = x0 + 2x1 >= 2x1 = [[a!6220!6220add#(mark(_x0), mark(_x1))]] [[mark#(add(_x0, _x1))]] = x0 + 2x1 >= x0 = [[mark#(_x0)]] [[mark#(add(_x0, _x1))]] = x0 + 2x1 >= x1 = [[mark#(_x1)]] [[mark#(dbl(_x0))]] = x0 >= x0 = [[mark#(_x0)]] [[mark#(first(_x0, _x1))]] = x0 + 2x1 >= 2x1 = [[a!6220!6220first#(mark(_x0), mark(_x1))]] [[mark#(first(_x0, _x1))]] = x0 + 2x1 >= x0 = [[mark#(_x0)]] [[mark#(first(_x0, _x1))]] = x0 + 2x1 >= x1 = [[mark#(_x1)]] [[mark#(cons(_x0, _x1))]] = 1 + x0 > x0 = [[mark#(_x0)]] [[mark#(recip(_x0))]] = x0 >= x0 = [[mark#(_x0)]] [[a!6220!6220terms(_x0)]] = 2 + 2x0 >= 1 + 2x0 = [[cons(recip(a!6220!6220sqr(mark(_x0))), terms(s(_x0)))]] [[a!6220!6220sqr(0)]] = 0 >= 0 = [[0]] [[a!6220!6220sqr(s(_x0))]] = 0 >= 0 = [[s(add(sqr(_x0), dbl(_x0)))]] [[a!6220!6220dbl(0)]] = 0 >= 0 = [[0]] [[a!6220!6220dbl(s(_x0))]] = 0 >= 0 = [[s(s(dbl(_x0)))]] [[a!6220!6220add(0, _x0)]] = 2x0 >= 2x0 = [[mark(_x0)]] [[a!6220!6220add(s(_x0), _x1)]] = 2x1 >= 0 = [[s(add(_x0, _x1))]] [[a!6220!6220first(s(_x0), cons(_x1, _x2))]] = 2 + 2x1 >= 1 + 2x1 = [[cons(mark(_x1), first(_x0, _x2))]] [[mark(terms(_x0))]] = 4 + 4x0 >= 2 + 4x0 = [[a!6220!6220terms(mark(_x0))]] [[mark(sqr(_x0))]] = 2x0 >= 2x0 = [[a!6220!6220sqr(mark(_x0))]] [[mark(add(_x0, _x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[a!6220!6220add(mark(_x0), mark(_x1))]] [[mark(dbl(_x0))]] = 2x0 >= 2x0 = [[a!6220!6220dbl(mark(_x0))]] [[mark(first(_x0, _x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[a!6220!6220first(mark(_x0), mark(_x1))]] [[mark(cons(_x0, _x1))]] = 2 + 2x0 >= 1 + 2x0 = [[cons(mark(_x0), _x1)]] [[mark(recip(_x0))]] = 2x0 >= 2x0 = [[recip(mark(_x0))]] [[mark(s(_x0))]] = 0 >= 0 = [[s(_x0)]] [[mark(0)]] = 0 >= 0 = [[0]] [[a!6220!6220terms(_x0)]] = 2 + 2x0 >= 2 + 2x0 = [[terms(_x0)]] [[a!6220!6220sqr(_x0)]] = x0 >= x0 = [[sqr(_x0)]] [[a!6220!6220add(_x0, _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[add(_x0, _x1)]] [[a!6220!6220dbl(_x0)]] = x0 >= x0 = [[dbl(_x0)]] [[a!6220!6220first(_x0, _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[first(_x0, _x1)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_3, R_1, minimal, formative) by (P_4, R_1, minimal, formative), where P_4 consists of: a!6220!6220add#(0, X) =#> mark#(X) mark#(sqr(X)) =#> mark#(X) mark#(add(X, Y)) =#> a!6220!6220add#(mark(X), mark(Y)) mark#(add(X, Y)) =#> mark#(X) mark#(add(X, Y)) =#> mark#(Y) mark#(dbl(X)) =#> mark#(X) mark#(first(X, Y)) =#> a!6220!6220first#(mark(X), mark(Y)) mark#(first(X, Y)) =#> mark#(X) mark#(first(X, Y)) =#> mark#(Y) mark#(recip(X)) =#> mark#(X) Thus, the original system is terminating if (P_4, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_4, R_1, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : 1, 2, 3, 4, 5, 6, 7, 8, 9 * 1 : 1, 2, 3, 4, 5, 6, 7, 8, 9 * 2 : 0 * 3 : 1, 2, 3, 4, 5, 6, 7, 8, 9 * 4 : 1, 2, 3, 4, 5, 6, 7, 8, 9 * 5 : 1, 2, 3, 4, 5, 6, 7, 8, 9 * 6 : * 7 : 1, 2, 3, 4, 5, 6, 7, 8, 9 * 8 : 1, 2, 3, 4, 5, 6, 7, 8, 9 * 9 : 1, 2, 3, 4, 5, 6, 7, 8, 9 This graph has the following strongly connected components: P_5: a!6220!6220add#(0, X) =#> mark#(X) mark#(sqr(X)) =#> mark#(X) mark#(add(X, Y)) =#> a!6220!6220add#(mark(X), mark(Y)) mark#(add(X, Y)) =#> mark#(X) mark#(add(X, Y)) =#> mark#(Y) mark#(dbl(X)) =#> mark#(X) mark#(first(X, Y)) =#> mark#(X) mark#(first(X, Y)) =#> mark#(Y) mark#(recip(X)) =#> mark#(X) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_4, R_1, m, f) by (P_5, R_1, m, f). Thus, the original system is terminating if (P_5, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_5, R_1, minimal, formative). The formative rules of (P_5, R_1) are R_2 ::= a!6220!6220sqr(0) => 0 a!6220!6220dbl(0) => 0 a!6220!6220add(0, X) => mark(X) mark(sqr(X)) => a!6220!6220sqr(mark(X)) mark(add(X, Y)) => a!6220!6220add(mark(X), mark(Y)) mark(dbl(X)) => a!6220!6220dbl(mark(X)) mark(first(X, Y)) => a!6220!6220first(mark(X), mark(Y)) mark(recip(X)) => recip(mark(X)) mark(0) => 0 a!6220!6220sqr(X) => sqr(X) a!6220!6220add(X, Y) => add(X, Y) a!6220!6220dbl(X) => dbl(X) a!6220!6220first(X, Y) => first(X, Y) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_5, R_1, minimal, formative) by (P_5, R_2, minimal, formative). Thus, the original system is terminating if (P_5, R_2, minimal, formative) is finite. We consider the dependency pair problem (P_5, R_2, minimal, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: a!6220!6220add#(0, X) >? mark#(X) mark#(sqr(X)) >? mark#(X) mark#(add(X, Y)) >? a!6220!6220add#(mark(X), mark(Y)) mark#(add(X, Y)) >? mark#(X) mark#(add(X, Y)) >? mark#(Y) mark#(dbl(X)) >? mark#(X) mark#(first(X, Y)) >? mark#(X) mark#(first(X, Y)) >? mark#(Y) mark#(recip(X)) >? mark#(X) a!6220!6220sqr(0) >= 0 a!6220!6220dbl(0) >= 0 a!6220!6220add(0, X) >= mark(X) mark(sqr(X)) >= a!6220!6220sqr(mark(X)) mark(add(X, Y)) >= a!6220!6220add(mark(X), mark(Y)) mark(dbl(X)) >= a!6220!6220dbl(mark(X)) mark(first(X, Y)) >= a!6220!6220first(mark(X), mark(Y)) mark(recip(X)) >= recip(mark(X)) mark(0) >= 0 a!6220!6220sqr(X) >= sqr(X) a!6220!6220add(X, Y) >= add(X, Y) a!6220!6220dbl(X) >= dbl(X) a!6220!6220first(X, Y) >= first(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 1 a!6220!6220add = \y0y1.2y0 + 2y1 a!6220!6220add# = \y0y1.2y0 + 2y1 a!6220!6220dbl = \y0.y0 a!6220!6220first = \y0y1.y0 + 2y1 a!6220!6220sqr = \y0.y0 add = \y0y1.2y0 + 2y1 dbl = \y0.y0 first = \y0y1.y0 + 2y1 mark = \y0.2y0 mark# = \y0.2y0 recip = \y0.2y0 sqr = \y0.y0 Using this interpretation, the requirements translate to: [[a!6220!6220add#(0, _x0)]] = 2 + 2x0 > 2x0 = [[mark#(_x0)]] [[mark#(sqr(_x0))]] = 2x0 >= 2x0 = [[mark#(_x0)]] [[mark#(add(_x0, _x1))]] = 4x0 + 4x1 >= 4x0 + 4x1 = [[a!6220!6220add#(mark(_x0), mark(_x1))]] [[mark#(add(_x0, _x1))]] = 4x0 + 4x1 >= 2x0 = [[mark#(_x0)]] [[mark#(add(_x0, _x1))]] = 4x0 + 4x1 >= 2x1 = [[mark#(_x1)]] [[mark#(dbl(_x0))]] = 2x0 >= 2x0 = [[mark#(_x0)]] [[mark#(first(_x0, _x1))]] = 2x0 + 4x1 >= 2x0 = [[mark#(_x0)]] [[mark#(first(_x0, _x1))]] = 2x0 + 4x1 >= 2x1 = [[mark#(_x1)]] [[mark#(recip(_x0))]] = 4x0 >= 2x0 = [[mark#(_x0)]] [[a!6220!6220sqr(0)]] = 1 >= 1 = [[0]] [[a!6220!6220dbl(0)]] = 1 >= 1 = [[0]] [[a!6220!6220add(0, _x0)]] = 2 + 2x0 >= 2x0 = [[mark(_x0)]] [[mark(sqr(_x0))]] = 2x0 >= 2x0 = [[a!6220!6220sqr(mark(_x0))]] [[mark(add(_x0, _x1))]] = 4x0 + 4x1 >= 4x0 + 4x1 = [[a!6220!6220add(mark(_x0), mark(_x1))]] [[mark(dbl(_x0))]] = 2x0 >= 2x0 = [[a!6220!6220dbl(mark(_x0))]] [[mark(first(_x0, _x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[a!6220!6220first(mark(_x0), mark(_x1))]] [[mark(recip(_x0))]] = 4x0 >= 4x0 = [[recip(mark(_x0))]] [[mark(0)]] = 2 >= 1 = [[0]] [[a!6220!6220sqr(_x0)]] = x0 >= x0 = [[sqr(_x0)]] [[a!6220!6220add(_x0, _x1)]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[add(_x0, _x1)]] [[a!6220!6220dbl(_x0)]] = x0 >= x0 = [[dbl(_x0)]] [[a!6220!6220first(_x0, _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[first(_x0, _x1)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_5, R_2, minimal, formative) by (P_6, R_2, minimal, formative), where P_6 consists of: mark#(sqr(X)) =#> mark#(X) mark#(add(X, Y)) =#> a!6220!6220add#(mark(X), mark(Y)) mark#(add(X, Y)) =#> mark#(X) mark#(add(X, Y)) =#> mark#(Y) mark#(dbl(X)) =#> mark#(X) mark#(first(X, Y)) =#> mark#(X) mark#(first(X, Y)) =#> mark#(Y) mark#(recip(X)) =#> mark#(X) Thus, the original system is terminating if (P_6, R_2, minimal, formative) is finite. We consider the dependency pair problem (P_6, R_2, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : 0, 1, 2, 3, 4, 5, 6, 7 * 1 : * 2 : 0, 1, 2, 3, 4, 5, 6, 7 * 3 : 0, 1, 2, 3, 4, 5, 6, 7 * 4 : 0, 1, 2, 3, 4, 5, 6, 7 * 5 : 0, 1, 2, 3, 4, 5, 6, 7 * 6 : 0, 1, 2, 3, 4, 5, 6, 7 * 7 : 0, 1, 2, 3, 4, 5, 6, 7 This graph has the following strongly connected components: P_7: mark#(sqr(X)) =#> mark#(X) mark#(add(X, Y)) =#> mark#(X) mark#(add(X, Y)) =#> mark#(Y) mark#(dbl(X)) =#> mark#(X) mark#(first(X, Y)) =#> mark#(X) mark#(first(X, Y)) =#> mark#(Y) mark#(recip(X)) =#> mark#(X) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_6, R_2, m, f) by (P_7, R_2, m, f). Thus, the original system is terminating if (P_7, R_2, minimal, formative) is finite. We consider the dependency pair problem (P_7, R_2, minimal, formative). We apply the subterm criterion with the following projection function: nu(mark#) = 1 Thus, we can orient the dependency pairs as follows: nu(mark#(sqr(X))) = sqr(X) |> X = nu(mark#(X)) nu(mark#(add(X, Y))) = add(X, Y) |> X = nu(mark#(X)) nu(mark#(add(X, Y))) = add(X, Y) |> Y = nu(mark#(Y)) nu(mark#(dbl(X))) = dbl(X) |> X = nu(mark#(X)) nu(mark#(first(X, Y))) = first(X, Y) |> X = nu(mark#(X)) nu(mark#(first(X, Y))) = first(X, Y) |> Y = nu(mark#(Y)) nu(mark#(recip(X))) = recip(X) |> X = nu(mark#(X)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_7, R_2, minimal, f) by ({}, R_2, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. [KusIsoSakBla09] K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui. Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems. In volume 92(10) of IEICE Transactions on Information and Systems. 2007--2015, 2009.