details

property	value
status	complete
benchmark	filliatre2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n104.star.cs.uiowa.edu
space	CiME_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.33318400383 seconds
cpu usage	0.298858491
max memory	5885952.0

stage attributes

key	value
output-size	15791
starexec-result	YES

output

/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 : [] --> o B : [] --> o C : [] --> o f : [o * o] --> o f!450 : [o * o] --> o f!450!450 : [o] --> o foldB : [o * o] --> o foldC : [o * o] --> o g : [o] --> o s : [o] --> o triple : [o * o * o] --> o g(A) => A g(B) => A g(B) => B g(C) => A g(C) => B g(C) => C foldB(X, 0) => X foldB(X, s(Y)) => f(foldB(X, Y), B) foldC(X, 0) => X foldC(X, s(Y)) => f(foldC(X, Y), C) f(X, Y) => f!450(X, g(Y)) f!450(triple(X, Y, Z), C) => triple(X, Y, s(Z)) f!450(triple(X, Y, Z), B) => f(triple(X, Y, Z), A) f!450(triple(X, Y, Z), A) => f!450!450(foldB(triple(s(X), 0, Z), Y)) f!450!450(triple(X, Y, Z)) => foldC(triple(X, Y, 0), Z) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): g(A) >? A g(B) >? A g(B) >? B g(C) >? A g(C) >? B g(C) >? C foldB(X, 0) >? X foldB(X, s(Y)) >? f(foldB(X, Y), B) foldC(X, 0) >? X foldC(X, s(Y)) >? f(foldC(X, Y), C) f(X, Y) >? f!450(X, g(Y)) f!450(triple(X, Y, Z), C) >? triple(X, Y, s(Z)) f!450(triple(X, Y, Z), B) >? f(triple(X, Y, Z), A) f!450(triple(X, Y, Z), A) >? f!450!450(foldB(triple(s(X), 0, Z), Y)) f!450!450(triple(X, Y, Z)) >? foldC(triple(X, Y, 0), Z) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 A = 1 B = 1 C = 1 f = \y0y1.1 + y0 + y1 f!450 = \y0y1.1 + y0 + y1 f!450!450 = \y0.y0 foldB = \y0y1.y0 + 3y1 foldC = \y0y1.y0 + 2y1 g = \y0.y0 s = \y0.1 + y0 triple = \y0y1y2.2y0 + 2y2 + 3y1 Using this interpretation, the requirements translate to: [[g(A)]] = 1 >= 1 = [[A]] [[g(B)]] = 1 >= 1 = [[A]] [[g(B)]] = 1 >= 1 = [[B]] [[g(C)]] = 1 >= 1 = [[A]] [[g(C)]] = 1 >= 1 = [[B]] [[g(C)]] = 1 >= 1 = [[C]] [[foldB(_x0, 0)]] = x0 >= x0 = [[_x0]] [[foldB(_x0, s(_x1))]] = 3 + x0 + 3x1 > 2 + x0 + 3x1 = [[f(foldB(_x0, _x1), B)]] [[foldC(_x0, 0)]] = x0 >= x0 = [[_x0]] [[foldC(_x0, s(_x1))]] = 2 + x0 + 2x1 >= 2 + x0 + 2x1 = [[f(foldC(_x0, _x1), C)]] [[f(_x0, _x1)]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[f!450(_x0, g(_x1))]] [[f!450(triple(_x0, _x1, _x2), C)]] = 2 + 2x0 + 2x2 + 3x1 >= 2 + 2x0 + 2x2 + 3x1 = [[triple(_x0, _x1, s(_x2))]] [[f!450(triple(_x0, _x1, _x2), B)]] = 2 + 2x0 + 2x2 + 3x1 >= 2 + 2x0 + 2x2 + 3x1 = [[f(triple(_x0, _x1, _x2), A)]] [[f!450(triple(_x0, _x1, _x2), A)]] = 2 + 2x0 + 2x2 + 3x1 >= 2 + 2x0 + 2x2 + 3x1 = [[f!450!450(foldB(triple(s(_x0), 0, _x2), _x1))]] [[f!450!450(triple(_x0, _x1, _x2))]] = 2x0 + 2x2 + 3x1 >= 2x0 + 2x2 + 3x1 = [[foldC(triple(_x0, _x1, 0), _x2)]] We can thus remove the following rules: foldB(X, s(Y)) => f(foldB(X, Y), B) We use rule removal, following [Kop12, Theorem 2.23]. popout

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