details

property	value
status	complete
benchmark	06.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n027.star.cs.uiowa.edu
space	Various_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.0334310531616 seconds
cpu usage	0.029719484
max memory	1388544.0

stage attributes

key	value
output-size	2964
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. f : [o * o] --> o g1 : [o * o * o] --> o g2 : [o * o * o] --> o h : [o * o] --> o f(X, Y) => g1(X, X, Y) f(X, Y) => g1(Y, X, X) f(X, Y) => g2(X, Y, Y) f(X, Y) => g2(Y, Y, X) g1(X, X, Y) => h(X, Y) g1(X, Y, Y) => h(Y, X) g2(X, Y, Y) => h(X, Y) g2(X, X, Y) => h(Y, X) h(X, 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]): f(X, Y) >? g1(X, X, Y) f(X, Y) >? g1(Y, X, X) f(X, Y) >? g2(X, Y, Y) f(X, Y) >? g2(Y, Y, X) g1(X, X, Y) >? h(X, Y) g1(X, Y, Y) >? h(Y, X) g2(X, Y, Y) >? h(X, Y) g2(X, X, Y) >? h(Y, X) h(X, X) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: f = \y0y1.3 + 3y0 + 3y1 g1 = \y0y1y2.2 + y0 + y1 + y2 g2 = \y0y1y2.2 + y0 + y1 + y2 h = \y0y1.y0 + y1 Using this interpretation, the requirements translate to: [[f(_x0, _x1)]] = 3 + 3x0 + 3x1 > 2 + x1 + 2x0 = [[g1(_x0, _x0, _x1)]] [[f(_x0, _x1)]] = 3 + 3x0 + 3x1 > 2 + x1 + 2x0 = [[g1(_x1, _x0, _x0)]] [[f(_x0, _x1)]] = 3 + 3x0 + 3x1 > 2 + x0 + 2x1 = [[g2(_x0, _x1, _x1)]] [[f(_x0, _x1)]] = 3 + 3x0 + 3x1 > 2 + x0 + 2x1 = [[g2(_x1, _x1, _x0)]] [[g1(_x0, _x0, _x1)]] = 2 + x1 + 2x0 > x0 + x1 = [[h(_x0, _x1)]] [[g1(_x0, _x1, _x1)]] = 2 + x0 + 2x1 > x0 + x1 = [[h(_x1, _x0)]] [[g2(_x0, _x1, _x1)]] = 2 + x0 + 2x1 > x0 + x1 = [[h(_x0, _x1)]] [[g2(_x0, _x0, _x1)]] = 2 + x1 + 2x0 > x0 + x1 = [[h(_x1, _x0)]] [[h(_x0, _x0)]] = 2x0 >= x0 = [[_x0]] We can thus remove the following rules: f(X, Y) => g1(X, X, Y) f(X, Y) => g1(Y, X, X) f(X, Y) => g2(X, Y, Y) f(X, Y) => g2(Y, Y, X) g1(X, X, Y) => h(X, Y) g1(X, Y, Y) => h(Y, X) g2(X, Y, Y) => h(X, Y) g2(X, X, Y) => h(Y, 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]): h(X, X) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: h = \y0y1.1 + y0 + y1 Using this interpretation, the requirements translate to: [[h(_x0, _x0)]] = 1 + 2x0 > x0 = [[_x0]] We can thus remove the following rules: h(X, 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. popout

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