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TRS Stand 20472 pair #381712633
details
property
value
status
complete
benchmark
mfp95.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n101.star.cs.uiowa.edu
space
Rubio_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0217640399933 seconds
cpu usage
0.018641168
max memory
1396736.0
stage attributes
key
value
output-size
1291
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. f : [o * o] --> o h : [o] --> o s : [o] --> o f(s(X), Y) => h(s(f(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]): f(s(X), Y) >? h(s(f(h(Y), X))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: f = \y0y1.2y0 + 2y1 h = \y0.y0 s = \y0.2 + y0 Using this interpretation, the requirements translate to: [[f(s(_x0), _x1)]] = 4 + 2x0 + 2x1 > 2 + 2x0 + 2x1 = [[h(s(f(h(_x1), _x0)))]] We can thus remove the following rules: f(s(X), Y) => h(s(f(h(Y), 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.
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