details

property	value
status	complete
benchmark	bn122.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n013.star.cs.uiowa.edu
space	Rubio_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.0334579944611 seconds
cpu usage	0.027450279
max memory	1519616.0

stage attributes

key	value
output-size	2158
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. plus : [o * o] --> o s : [o] --> o times : [o * o] --> o plus(plus(X, Y), Z) => plus(X, plus(Y, Z)) times(X, s(Y)) => plus(X, times(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]): plus(plus(X, Y), Z) >? plus(X, plus(Y, Z)) times(X, s(Y)) >? plus(X, times(Y, X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: plus = \y0y1.y0 + y1 s = \y0.3 + 3y0 times = \y0y1.y1 + 2y0 Using this interpretation, the requirements translate to: [[plus(plus(_x0, _x1), _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[plus(_x0, plus(_x1, _x2))]] [[times(_x0, s(_x1))]] = 3 + 2x0 + 3x1 > 2x0 + 2x1 = [[plus(_x0, times(_x1, _x0))]] We can thus remove the following rules: times(X, s(Y)) => plus(X, times(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]): 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: plus = \y0y1.3 + y1 + 3y0 Using this interpretation, the requirements translate to: [[plus(plus(_x0, _x1), _x2)]] = 12 + x2 + 3x1 + 9x0 > 6 + x2 + 3x0 + 3x1 = [[plus(_x0, plus(_x1, _x2))]] We can thus remove the following rules: plus(plus(X, Y), Z) => plus(X, plus(Y, Z)) 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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