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TRS Stand 20472 pair #381714953
details
property
value
status
complete
benchmark
2.61.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n004.star.cs.uiowa.edu
space
SK90
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0617280006409 seconds
cpu usage
0.052684916
max memory
2129920.0
stage attributes
key
value
output-size
3819
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 f : [o * o] --> o g : [o] --> o h : [o] --> o h1 : [o * o] --> o h2 : [o * o * o] --> o i : [o] --> o j : [o * o] --> o k : [o] --> o s : [o] --> o f(j(X, Y), Y) => g(f(X, k(Y))) f(X, h1(Y, Z)) => h2(0, X, h1(Y, Z)) g(h2(X, Y, h1(Z, U))) => h2(s(X), Y, h1(Z, U)) h2(X, j(Y, h1(Z, U)), h1(Z, U)) => h2(s(X), Y, h1(s(Z), U)) i(f(X, h(Y))) => Y i(h2(s(X), Y, h1(X, Z))) => Z k(h(X)) => h1(0, X) k(h1(X, Y)) => h1(s(X), Y) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): f(j(X, Y), Y) >? g(f(X, k(Y))) f(X, h1(Y, Z)) >? h2(0, X, h1(Y, Z)) g(h2(X, Y, h1(Z, U))) >? h2(s(X), Y, h1(Z, U)) h2(X, j(Y, h1(Z, U)), h1(Z, U)) >? h2(s(X), Y, h1(s(Z), U)) i(f(X, h(Y))) >? Y i(h2(s(X), Y, h1(X, Z))) >? Z k(h(X)) >? h1(0, X) k(h1(X, Y)) >? h1(s(X), Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 f = \y0y1.1 + 2y1 + 3y0 g = \y0.2y0 h = \y0.3 + 3y0 h1 = \y0y1.y1 + 2y0 h2 = \y0y1y2.2y0 + 2y1 + 2y2 i = \y0.3 + y0 j = \y0y1.3 + 3y0 + 3y1 k = \y0.2 + y0 s = \y0.y0 Using this interpretation, the requirements translate to: [[f(j(_x0, _x1), _x1)]] = 10 + 9x0 + 11x1 >= 10 + 4x1 + 6x0 = [[g(f(_x0, k(_x1)))]] [[f(_x0, h1(_x1, _x2))]] = 1 + 2x2 + 3x0 + 4x1 > 2x0 + 2x2 + 4x1 = [[h2(0, _x0, h1(_x1, _x2))]] [[g(h2(_x0, _x1, h1(_x2, _x3)))]] = 4x0 + 4x1 + 4x3 + 8x2 >= 2x0 + 2x1 + 2x3 + 4x2 = [[h2(s(_x0), _x1, h1(_x2, _x3))]] [[h2(_x0, j(_x1, h1(_x2, _x3)), h1(_x2, _x3))]] = 6 + 2x0 + 6x1 + 8x3 + 16x2 > 2x0 + 2x1 + 2x3 + 4x2 = [[h2(s(_x0), _x1, h1(s(_x2), _x3))]] [[i(f(_x0, h(_x1)))]] = 10 + 3x0 + 6x1 > x1 = [[_x1]] [[i(h2(s(_x0), _x1, h1(_x0, _x2)))]] = 3 + 2x1 + 2x2 + 6x0 > x2 = [[_x2]] [[k(h(_x0))]] = 5 + 3x0 > x0 = [[h1(0, _x0)]] [[k(h1(_x0, _x1))]] = 2 + x1 + 2x0 > x1 + 2x0 = [[h1(s(_x0), _x1)]] We can thus remove the following rules: f(X, h1(Y, Z)) => h2(0, X, h1(Y, Z)) h2(X, j(Y, h1(Z, U)), h1(Z, U)) => h2(s(X), Y, h1(s(Z), U)) i(f(X, h(Y))) => Y i(h2(s(X), Y, h1(X, Z))) => Z k(h(X)) => h1(0, X) k(h1(X, Y)) => h1(s(X), Y) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): f(j(X, Y), Y) >? g(f(X, k(Y))) g(h2(X, Y, h1(Z, U))) >? h2(s(X), Y, h1(Z, U)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: f = \y0y1.y1 + 3y0 g = \y0.2y0 h1 = \y0y1.2 + 2y0 + 2y1 h2 = \y0y1y2.y0 + 2y1 + 2y2 j = \y0y1.3 + 3y0 + 3y1 k = \y0.y0 s = \y0.y0 Using this interpretation, the requirements translate to:
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