/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. !940 : [] --> o !minus : [o * o] --> o !plus : [o * o] --> o 0 : [o] --> o 1 : [o] --> o false : [] --> o ge : [o * o] --> o if : [o * o * o] --> o log : [o] --> o log!450 : [o] --> o not : [o] --> o true : [] --> o 0(!940) => !940 !plus(!940, X) => X !plus(X, !940) => X !plus(0(X), 0(Y)) => 0(!plus(X, Y)) !plus(0(X), 1(Y)) => 1(!plus(X, Y)) !plus(1(X), 0(Y)) => 1(!plus(X, Y)) !plus(1(X), 1(Y)) => 0(!plus(!plus(X, Y), 1(!940))) !plus(!plus(X, Y), Z) => !plus(X, !plus(Y, Z)) !minus(!940, X) => !940 !minus(X, !940) => X !minus(0(X), 0(Y)) => 0(!minus(X, Y)) !minus(0(X), 1(Y)) => 1(!minus(!minus(X, Y), 1(!940))) !minus(1(X), 0(Y)) => 1(!minus(X, Y)) !minus(1(X), 1(Y)) => 0(!minus(X, Y)) not(true) => false not(false) => true if(true, X, Y) => X if(false, X, Y) => Y ge(0(X), 0(Y)) => ge(X, Y) ge(0(X), 1(Y)) => not(ge(Y, X)) ge(1(X), 0(Y)) => ge(X, Y) ge(1(X), 1(Y)) => ge(X, Y) ge(X, !940) => true ge(!940, 0(X)) => ge(!940, X) ge(!940, 1(X)) => false log(X) => !minus(log!450(X), 1(!940)) log!450(!940) => !940 log!450(1(X)) => !plus(log!450(X), 1(!940)) log!450(0(X)) => if(ge(X, 1(!940)), !plus(log!450(X), 1(!940)), !940) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0(!940) >? !940 !plus(!940, X) >? X !plus(X, !940) >? X !plus(0(X), 0(Y)) >? 0(!plus(X, Y)) !plus(0(X), 1(Y)) >? 1(!plus(X, Y)) !plus(1(X), 0(Y)) >? 1(!plus(X, Y)) !plus(1(X), 1(Y)) >? 0(!plus(!plus(X, Y), 1(!940))) !plus(!plus(X, Y), Z) >? !plus(X, !plus(Y, Z)) !minus(!940, X) >? !940 !minus(X, !940) >? X !minus(0(X), 0(Y)) >? 0(!minus(X, Y)) !minus(0(X), 1(Y)) >? 1(!minus(!minus(X, Y), 1(!940))) !minus(1(X), 0(Y)) >? 1(!minus(X, Y)) !minus(1(X), 1(Y)) >? 0(!minus(X, Y)) not(true) >? false not(false) >? true if(true, X, Y) >? X if(false, X, Y) >? Y ge(0(X), 0(Y)) >? ge(X, Y) ge(0(X), 1(Y)) >? not(ge(Y, X)) ge(1(X), 0(Y)) >? ge(X, Y) ge(1(X), 1(Y)) >? ge(X, Y) ge(X, !940) >? true ge(!940, 0(X)) >? ge(!940, X) ge(!940, 1(X)) >? false log(X) >? !minus(log!450(X), 1(!940)) log!450(!940) >? !940 log!450(1(X)) >? !plus(log!450(X), 1(!940)) log!450(0(X)) >? if(ge(X, 1(!940)), !plus(log!450(X), 1(!940)), !940) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !940 = 0 !minus = \y0y1.y0 + y1 !plus = \y0y1.y0 + y1 0 = \y0.2y0 1 = \y0.2y0 false = 0 ge = \y0y1.y0 + y1 if = \y0y1y2.y0 + y1 + y2 log = \y0.3 + 3y0 log!450 = \y0.y0 not = \y0.y0 true = 0 Using this interpretation, the requirements translate to: [[0(!940)]] = 0 >= 0 = [[!940]] [[!plus(!940, _x0)]] = x0 >= x0 = [[_x0]] [[!plus(_x0, !940)]] = x0 >= x0 = [[_x0]] [[!plus(0(_x0), 0(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[0(!plus(_x0, _x1))]] [[!plus(0(_x0), 1(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[1(!plus(_x0, _x1))]] [[!plus(1(_x0), 0(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[1(!plus(_x0, _x1))]] [[!plus(1(_x0), 1(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[0(!plus(!plus(_x0, _x1), 1(!940)))]] [[!plus(!plus(_x0, _x1), _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[!plus(_x0, !plus(_x1, _x2))]] [[!minus(!940, _x0)]] = x0 >= 0 = [[!940]] [[!minus(_x0, !940)]] = x0 >= x0 = [[_x0]] [[!minus(0(_x0), 0(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[0(!minus(_x0, _x1))]] [[!minus(0(_x0), 1(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[1(!minus(!minus(_x0, _x1), 1(!940)))]] [[!minus(1(_x0), 0(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[1(!minus(_x0, _x1))]] [[!minus(1(_x0), 1(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[0(!minus(_x0, _x1))]] [[not(true)]] = 0 >= 0 = [[false]] [[not(false)]] = 0 >= 0 = [[true]] [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] [[if(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] [[ge(0(_x0), 0(_x1))]] = 2x0 + 2x1 >= x0 + x1 = [[ge(_x0, _x1)]] [[ge(0(_x0), 1(_x1))]] = 2x0 + 2x1 >= x0 + x1 = [[not(ge(_x1, _x0))]] [[ge(1(_x0), 0(_x1))]] = 2x0 + 2x1 >= x0 + x1 = [[ge(_x0, _x1)]] [[ge(1(_x0), 1(_x1))]] = 2x0 + 2x1 >= x0 + x1 = [[ge(_x0, _x1)]] [[ge(_x0, !940)]] = x0 >= 0 = [[true]] [[ge(!940, 0(_x0))]] = 2x0 >= x0 = [[ge(!940, _x0)]] [[ge(!940, 1(_x0))]] = 2x0 >= 0 = [[false]] [[log(_x0)]] = 3 + 3x0 > x0 = [[!minus(log!450(_x0), 1(!940))]] [[log!450(!940)]] = 0 >= 0 = [[!940]] [[log!450(1(_x0))]] = 2x0 >= x0 = [[!plus(log!450(_x0), 1(!940))]] [[log!450(0(_x0))]] = 2x0 >= 2x0 = [[if(ge(_x0, 1(!940)), !plus(log!450(_x0), 1(!940)), !940)]] We can thus remove the following rules: log(X) => !minus(log!450(X), 1(!940)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0(!940) >? !940 !plus(!940, X) >? X !plus(X, !940) >? X !plus(0(X), 0(Y)) >? 0(!plus(X, Y)) !plus(0(X), 1(Y)) >? 1(!plus(X, Y)) !plus(1(X), 0(Y)) >? 1(!plus(X, Y)) !plus(1(X), 1(Y)) >? 0(!plus(!plus(X, Y), 1(!940))) !plus(!plus(X, Y), Z) >? !plus(X, !plus(Y, Z)) !minus(!940, X) >? !940 !minus(X, !940) >? X !minus(0(X), 0(Y)) >? 0(!minus(X, Y)) !minus(0(X), 1(Y)) >? 1(!minus(!minus(X, Y), 1(!940))) !minus(1(X), 0(Y)) >? 1(!minus(X, Y)) !minus(1(X), 1(Y)) >? 0(!minus(X, Y)) not(true) >? false not(false) >? true if(true, X, Y) >? X if(false, X, Y) >? Y ge(0(X), 0(Y)) >? ge(X, Y) ge(0(X), 1(Y)) >? not(ge(Y, X)) ge(1(X), 0(Y)) >? ge(X, Y) ge(1(X), 1(Y)) >? ge(X, Y) ge(X, !940) >? true ge(!940, 0(X)) >? ge(!940, X) ge(!940, 1(X)) >? false log!450(!940) >? !940 log!450(1(X)) >? !plus(log!450(X), 1(!940)) log!450(0(X)) >? if(ge(X, 1(!940)), !plus(log!450(X), 1(!940)), !940) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !940 = 0 !minus = \y0y1.y0 + y1 !plus = \y0y1.y0 + y1 0 = \y0.2y0 1 = \y0.2y0 false = 0 ge = \y0y1.y0 + y1 if = \y0y1y2.y0 + y1 + y2 log!450 = \y0.1 + y0 not = \y0.2y0 true = 0 Using this interpretation, the requirements translate to: [[0(!940)]] = 0 >= 0 = [[!940]] [[!plus(!940, _x0)]] = x0 >= x0 = [[_x0]] [[!plus(_x0, !940)]] = x0 >= x0 = [[_x0]] [[!plus(0(_x0), 0(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[0(!plus(_x0, _x1))]] [[!plus(0(_x0), 1(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[1(!plus(_x0, _x1))]] [[!plus(1(_x0), 0(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[1(!plus(_x0, _x1))]] [[!plus(1(_x0), 1(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[0(!plus(!plus(_x0, _x1), 1(!940)))]] [[!plus(!plus(_x0, _x1), _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[!plus(_x0, !plus(_x1, _x2))]] [[!minus(!940, _x0)]] = x0 >= 0 = [[!940]] [[!minus(_x0, !940)]] = x0 >= x0 = [[_x0]] [[!minus(0(_x0), 0(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[0(!minus(_x0, _x1))]] [[!minus(0(_x0), 1(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[1(!minus(!minus(_x0, _x1), 1(!940)))]] [[!minus(1(_x0), 0(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[1(!minus(_x0, _x1))]] [[!minus(1(_x0), 1(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[0(!minus(_x0, _x1))]] [[not(true)]] = 0 >= 0 = [[false]] [[not(false)]] = 0 >= 0 = [[true]] [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] [[if(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] [[ge(0(_x0), 0(_x1))]] = 2x0 + 2x1 >= x0 + x1 = [[ge(_x0, _x1)]] [[ge(0(_x0), 1(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[not(ge(_x1, _x0))]] [[ge(1(_x0), 0(_x1))]] = 2x0 + 2x1 >= x0 + x1 = [[ge(_x0, _x1)]] [[ge(1(_x0), 1(_x1))]] = 2x0 + 2x1 >= x0 + x1 = [[ge(_x0, _x1)]] [[ge(_x0, !940)]] = x0 >= 0 = [[true]] [[ge(!940, 0(_x0))]] = 2x0 >= x0 = [[ge(!940, _x0)]] [[ge(!940, 1(_x0))]] = 2x0 >= 0 = [[false]] [[log!450(!940)]] = 1 > 0 = [[!940]] [[log!450(1(_x0))]] = 1 + 2x0 >= 1 + x0 = [[!plus(log!450(_x0), 1(!940))]] [[log!450(0(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[if(ge(_x0, 1(!940)), !plus(log!450(_x0), 1(!940)), !940)]] We can thus remove the following rules: log!450(!940) => !940 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] !plus#(0(X), 0(Y)) =#> 0#(!plus(X, Y)) 1] !plus#(0(X), 0(Y)) =#> !plus#(X, Y) 2] !plus#(0(X), 1(Y)) =#> !plus#(X, Y) 3] !plus#(1(X), 0(Y)) =#> !plus#(X, Y) 4] !plus#(1(X), 1(Y)) =#> 0#(!plus(!plus(X, Y), 1(!940))) 5] !plus#(1(X), 1(Y)) =#> !plus#(!plus(X, Y), 1(!940)) 6] !plus#(1(X), 1(Y)) =#> !plus#(X, Y) 7] !plus#(!plus(X, Y), Z) =#> !plus#(X, !plus(Y, Z)) 8] !plus#(!plus(X, Y), Z) =#> !plus#(Y, Z) 9] !minus#(0(X), 0(Y)) =#> 0#(!minus(X, Y)) 10] !minus#(0(X), 0(Y)) =#> !minus#(X, Y) 11] !minus#(0(X), 1(Y)) =#> !minus#(!minus(X, Y), 1(!940)) 12] !minus#(0(X), 1(Y)) =#> !minus#(X, Y) 13] !minus#(1(X), 0(Y)) =#> !minus#(X, Y) 14] !minus#(1(X), 1(Y)) =#> 0#(!minus(X, Y)) 15] !minus#(1(X), 1(Y)) =#> !minus#(X, Y) 16] ge#(0(X), 0(Y)) =#> ge#(X, Y) 17] ge#(0(X), 1(Y)) =#> not#(ge(Y, X)) 18] ge#(0(X), 1(Y)) =#> ge#(Y, X) 19] ge#(1(X), 0(Y)) =#> ge#(X, Y) 20] ge#(1(X), 1(Y)) =#> ge#(X, Y) 21] ge#(!940, 0(X)) =#> ge#(!940, X) 22] log!450#(1(X)) =#> !plus#(log!450(X), 1(!940)) 23] log!450#(1(X)) =#> log!450#(X) 24] log!450#(0(X)) =#> if#(ge(X, 1(!940)), !plus(log!450(X), 1(!940)), !940) 25] log!450#(0(X)) =#> ge#(X, 1(!940)) 26] log!450#(0(X)) =#> !plus#(log!450(X), 1(!940)) 27] log!450#(0(X)) =#> log!450#(X) Rules R_0: 0(!940) => !940 !plus(!940, X) => X !plus(X, !940) => X !plus(0(X), 0(Y)) => 0(!plus(X, Y)) !plus(0(X), 1(Y)) => 1(!plus(X, Y)) !plus(1(X), 0(Y)) => 1(!plus(X, Y)) !plus(1(X), 1(Y)) => 0(!plus(!plus(X, Y), 1(!940))) !plus(!plus(X, Y), Z) => !plus(X, !plus(Y, Z)) !minus(!940, X) => !940 !minus(X, !940) => X !minus(0(X), 0(Y)) => 0(!minus(X, Y)) !minus(0(X), 1(Y)) => 1(!minus(!minus(X, Y), 1(!940))) !minus(1(X), 0(Y)) => 1(!minus(X, Y)) !minus(1(X), 1(Y)) => 0(!minus(X, Y)) not(true) => false not(false) => true if(true, X, Y) => X if(false, X, Y) => Y ge(0(X), 0(Y)) => ge(X, Y) ge(0(X), 1(Y)) => not(ge(Y, X)) ge(1(X), 0(Y)) => ge(X, Y) ge(1(X), 1(Y)) => ge(X, Y) ge(X, !940) => true ge(!940, 0(X)) => ge(!940, X) ge(!940, 1(X)) => false log!450(1(X)) => !plus(log!450(X), 1(!940)) log!450(0(X)) => if(ge(X, 1(!940)), !plus(log!450(X), 1(!940)), !940) 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 : 0, 1, 2, 3, 4, 5, 6, 7, 8 * 2 : 0, 1, 2, 3, 4, 5, 6, 7, 8 * 3 : 0, 1, 2, 3, 4, 5, 6, 7, 8 * 4 : * 5 : 2, 4, 5, 6, 7, 8 * 6 : 0, 1, 2, 3, 4, 5, 6, 7, 8 * 7 : 0, 1, 2, 3, 4, 5, 6, 7, 8 * 8 : 0, 1, 2, 3, 4, 5, 6, 7, 8 * 9 : * 10 : 9, 10, 11, 12, 13, 14, 15 * 11 : 11, 12, 14, 15 * 12 : 9, 10, 11, 12, 13, 14, 15 * 13 : 9, 10, 11, 12, 13, 14, 15 * 14 : * 15 : 9, 10, 11, 12, 13, 14, 15 * 16 : 16, 17, 18, 19, 20, 21 * 17 : * 18 : 16, 17, 18, 19, 20, 21 * 19 : 16, 17, 18, 19, 20, 21 * 20 : 16, 17, 18, 19, 20, 21 * 21 : 21 * 22 : 2, 4, 5, 6, 7, 8 * 23 : 22, 23, 24, 25, 26, 27 * 24 : * 25 : 17, 18, 20 * 26 : 2, 4, 5, 6, 7, 8 * 27 : 22, 23, 24, 25, 26, 27 This graph has the following strongly connected components: P_1: !plus#(0(X), 0(Y)) =#> !plus#(X, Y) !plus#(0(X), 1(Y)) =#> !plus#(X, Y) !plus#(1(X), 0(Y)) =#> !plus#(X, Y) !plus#(1(X), 1(Y)) =#> !plus#(!plus(X, Y), 1(!940)) !plus#(1(X), 1(Y)) =#> !plus#(X, Y) !plus#(!plus(X, Y), Z) =#> !plus#(X, !plus(Y, Z)) !plus#(!plus(X, Y), Z) =#> !plus#(Y, Z) P_2: !minus#(0(X), 0(Y)) =#> !minus#(X, Y) !minus#(0(X), 1(Y)) =#> !minus#(!minus(X, Y), 1(!940)) !minus#(0(X), 1(Y)) =#> !minus#(X, Y) !minus#(1(X), 0(Y)) =#> !minus#(X, Y) !minus#(1(X), 1(Y)) =#> !minus#(X, Y) P_3: ge#(0(X), 0(Y)) =#> ge#(X, Y) ge#(0(X), 1(Y)) =#> ge#(Y, X) ge#(1(X), 0(Y)) =#> ge#(X, Y) ge#(1(X), 1(Y)) =#> ge#(X, Y) P_4: ge#(!940, 0(X)) =#> ge#(!940, X) P_5: log!450#(1(X)) =#> log!450#(X) log!450#(0(X)) =#> log!450#(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), (P_2, R_0, m, f), (P_3, R_0, m, f), (P_4, R_0, m, f) and (P_5, R_0, m, f). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative) and (P_5, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_5, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(log!450#) = 1 Thus, we can orient the dependency pairs as follows: nu(log!450#(1(X))) = 1(X) |> X = nu(log!450#(X)) nu(log!450#(0(X))) = 0(X) |> X = nu(log!450#(X)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_5, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative) and (P_4, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_4, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(ge#) = 2 Thus, we can orient the dependency pairs as follows: nu(ge#(!940, 0(X))) = 0(X) |> X = nu(ge#(!940, X)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_4, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_3, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_3, R_0, minimal, formative). We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. (P_3, R_0) has no usable rules. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: ge#(0(X), 0(Y)) >? ge#(X, Y) ge#(0(X), 1(Y)) >? ge#(Y, X) ge#(1(X), 0(Y)) >? ge#(X, Y) ge#(1(X), 1(Y)) >? ge#(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = \y0.3 + y0 1 = \y0.3 + y0 ge# = \y0y1.y0 + y1 Using this interpretation, the requirements translate to: [[ge#(0(_x0), 0(_x1))]] = 6 + x0 + x1 > x0 + x1 = [[ge#(_x0, _x1)]] [[ge#(0(_x0), 1(_x1))]] = 6 + x0 + x1 > x0 + x1 = [[ge#(_x1, _x0)]] [[ge#(1(_x0), 0(_x1))]] = 6 + x0 + x1 > x0 + x1 = [[ge#(_x0, _x1)]] [[ge#(1(_x0), 1(_x1))]] = 6 + x0 + x1 > x0 + x1 = [[ge#(_x0, _x1)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_3, R_0) by ({}, R_0). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_0, minimal, formative). We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_2, R_0) are: 0(!940) => !940 !minus(!940, X) => !940 !minus(X, !940) => X !minus(0(X), 0(Y)) => 0(!minus(X, Y)) !minus(0(X), 1(Y)) => 1(!minus(!minus(X, Y), 1(!940))) !minus(1(X), 0(Y)) => 1(!minus(X, Y)) !minus(1(X), 1(Y)) => 0(!minus(X, Y)) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: !minus#(0(X), 0(Y)) >? !minus#(X, Y) !minus#(0(X), 1(Y)) >? !minus#(!minus(X, Y), 1(!940)) !minus#(0(X), 1(Y)) >? !minus#(X, Y) !minus#(1(X), 0(Y)) >? !minus#(X, Y) !minus#(1(X), 1(Y)) >? !minus#(X, Y) 0(!940) >= !940 !minus(!940, X) >= !940 !minus(X, !940) >= X !minus(0(X), 0(Y)) >= 0(!minus(X, Y)) !minus(0(X), 1(Y)) >= 1(!minus(!minus(X, Y), 1(!940))) !minus(1(X), 0(Y)) >= 1(!minus(X, Y)) !minus(1(X), 1(Y)) >= 0(!minus(X, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !940 = 0 !minus = \y0y1.y0 !minus# = \y0y1.y0 0 = \y0.1 + 2y0 1 = \y0.1 + 2y0 Using this interpretation, the requirements translate to: [[!minus#(0(_x0), 0(_x1))]] = 1 + 2x0 > x0 = [[!minus#(_x0, _x1)]] [[!minus#(0(_x0), 1(_x1))]] = 1 + 2x0 > x0 = [[!minus#(!minus(_x0, _x1), 1(!940))]] [[!minus#(0(_x0), 1(_x1))]] = 1 + 2x0 > x0 = [[!minus#(_x0, _x1)]] [[!minus#(1(_x0), 0(_x1))]] = 1 + 2x0 > x0 = [[!minus#(_x0, _x1)]] [[!minus#(1(_x0), 1(_x1))]] = 1 + 2x0 > x0 = [[!minus#(_x0, _x1)]] [[0(!940)]] = 1 >= 0 = [[!940]] [[!minus(!940, _x0)]] = 0 >= 0 = [[!940]] [[!minus(_x0, !940)]] = x0 >= x0 = [[_x0]] [[!minus(0(_x0), 0(_x1))]] = 1 + 2x0 >= 1 + 2x0 = [[0(!minus(_x0, _x1))]] [[!minus(0(_x0), 1(_x1))]] = 1 + 2x0 >= 1 + 2x0 = [[1(!minus(!minus(_x0, _x1), 1(!940)))]] [[!minus(1(_x0), 0(_x1))]] = 1 + 2x0 >= 1 + 2x0 = [[1(!minus(_x0, _x1))]] [[!minus(1(_x0), 1(_x1))]] = 1 + 2x0 >= 1 + 2x0 = [[0(!minus(_x0, _x1))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, R_0) by ({}, R_0). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. 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). We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_1, R_0) are: 0(!940) => !940 !plus(!940, X) => X !plus(X, !940) => X !plus(0(X), 0(Y)) => 0(!plus(X, Y)) !plus(0(X), 1(Y)) => 1(!plus(X, Y)) !plus(1(X), 0(Y)) => 1(!plus(X, Y)) !plus(1(X), 1(Y)) => 0(!plus(!plus(X, Y), 1(!940))) !plus(!plus(X, Y), Z) => !plus(X, !plus(Y, Z)) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: !plus#(0(X), 0(Y)) >? !plus#(X, Y) !plus#(0(X), 1(Y)) >? !plus#(X, Y) !plus#(1(X), 0(Y)) >? !plus#(X, Y) !plus#(1(X), 1(Y)) >? !plus#(!plus(X, Y), 1(!940)) !plus#(1(X), 1(Y)) >? !plus#(X, Y) !plus#(!plus(X, Y), Z) >? !plus#(X, !plus(Y, Z)) !plus#(!plus(X, Y), Z) >? !plus#(Y, Z) 0(!940) >= !940 !plus(!940, X) >= X !plus(X, !940) >= X !plus(0(X), 0(Y)) >= 0(!plus(X, Y)) !plus(0(X), 1(Y)) >= 1(!plus(X, Y)) !plus(1(X), 0(Y)) >= 1(!plus(X, Y)) !plus(1(X), 1(Y)) >= 0(!plus(!plus(X, Y), 1(!940))) !plus(!plus(X, Y), Z) >= !plus(X, !plus(Y, Z)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !940 = 0 !plus = \y0y1.y0 + y1 !plus# = \y0y1.2y0 + 2y1 0 = \y0.2y0 1 = \y0.2 + 2y0 Using this interpretation, the requirements translate to: [[!plus#(0(_x0), 0(_x1))]] = 4x0 + 4x1 >= 2x0 + 2x1 = [[!plus#(_x0, _x1)]] [[!plus#(0(_x0), 1(_x1))]] = 4 + 4x0 + 4x1 > 2x0 + 2x1 = [[!plus#(_x0, _x1)]] [[!plus#(1(_x0), 0(_x1))]] = 4 + 4x0 + 4x1 > 2x0 + 2x1 = [[!plus#(_x0, _x1)]] [[!plus#(1(_x0), 1(_x1))]] = 8 + 4x0 + 4x1 > 4 + 2x0 + 2x1 = [[!plus#(!plus(_x0, _x1), 1(!940))]] [[!plus#(1(_x0), 1(_x1))]] = 8 + 4x0 + 4x1 > 2x0 + 2x1 = [[!plus#(_x0, _x1)]] [[!plus#(!plus(_x0, _x1), _x2)]] = 2x0 + 2x1 + 2x2 >= 2x0 + 2x1 + 2x2 = [[!plus#(_x0, !plus(_x1, _x2))]] [[!plus#(!plus(_x0, _x1), _x2)]] = 2x0 + 2x1 + 2x2 >= 2x1 + 2x2 = [[!plus#(_x1, _x2)]] [[0(!940)]] = 0 >= 0 = [[!940]] [[!plus(!940, _x0)]] = x0 >= x0 = [[_x0]] [[!plus(_x0, !940)]] = x0 >= x0 = [[_x0]] [[!plus(0(_x0), 0(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[0(!plus(_x0, _x1))]] [[!plus(0(_x0), 1(_x1))]] = 2 + 2x0 + 2x1 >= 2 + 2x0 + 2x1 = [[1(!plus(_x0, _x1))]] [[!plus(1(_x0), 0(_x1))]] = 2 + 2x0 + 2x1 >= 2 + 2x0 + 2x1 = [[1(!plus(_x0, _x1))]] [[!plus(1(_x0), 1(_x1))]] = 4 + 2x0 + 2x1 >= 4 + 2x0 + 2x1 = [[0(!plus(!plus(_x0, _x1), 1(!940)))]] [[!plus(!plus(_x0, _x1), _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[!plus(_x0, !plus(_x1, _x2))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_6, R_0, minimal, formative), where P_6 consists of: !plus#(0(X), 0(Y)) =#> !plus#(X, Y) !plus#(!plus(X, Y), Z) =#> !plus#(X, !plus(Y, Z)) !plus#(!plus(X, Y), Z) =#> !plus#(Y, Z) Thus, the original system is terminating if (P_6, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_6, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(!plus#) = 1 Thus, we can orient the dependency pairs as follows: nu(!plus#(0(X), 0(Y))) = 0(X) |> X = nu(!plus#(X, Y)) nu(!plus#(!plus(X, Y), Z)) = !plus(X, Y) |> X = nu(!plus#(X, !plus(Y, Z))) nu(!plus#(!plus(X, Y), Z)) = !plus(X, Y) |> Y = nu(!plus#(Y, Z)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_6, R_0, minimal, f) by ({}, R_0, 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.