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TRS Stand 20472 pair #381710067
details
property
value
status
complete
benchmark
t011.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n078.star.cs.uiowa.edu
space
HirokawaMiddeldorp_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0656340122223 seconds
cpu usage
0.026217701
max memory
1241088.0
stage attributes
key
value
output-size
1971
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 g : [o * o] --> o h : [o * o] --> o g(f(X), Y) => f(h(X, Y)) h(X, Y) => g(X, f(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]): g(f(X), Y) >? f(h(X, Y)) h(X, Y) >? g(X, f(Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: f = \y0.1 + y0 g = \y0y1.y1 + 3y0 h = \y0y1.1 + y1 + 3y0 Using this interpretation, the requirements translate to: [[g(f(_x0), _x1)]] = 3 + x1 + 3x0 > 2 + x1 + 3x0 = [[f(h(_x0, _x1))]] [[h(_x0, _x1)]] = 1 + x1 + 3x0 >= 1 + x1 + 3x0 = [[g(_x0, f(_x1))]] We can thus remove the following rules: g(f(X), Y) => f(h(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]): h(X, Y) >? g(X, f(Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: f = \y0.y0 g = \y0y1.y0 + y1 h = \y0y1.3 + 3y0 + 3y1 Using this interpretation, the requirements translate to: [[h(_x0, _x1)]] = 3 + 3x0 + 3x1 > x0 + x1 = [[g(_x0, f(_x1))]] We can thus remove the following rules: h(X, Y) => g(X, f(Y)) 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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