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TRS Stand 20472 pair #381710061
details
property
value
status
complete
benchmark
Ex2_6_1Composition.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n047.star.cs.uiowa.edu
space
Applicative_05
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0228309631348 seconds
cpu usage
0.019038541
max memory
1400832.0
stage attributes
key
value
output-size
1361
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. app : [o * o] --> o compose : [] --> o app(app(app(compose, X), Y), Z) => app(X, app(Y, Z)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): app(app(app(compose, X), Y), Z) >? app(X, app(Y, Z)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: app = \y0y1.y1 + 2y0 compose = 3 Using this interpretation, the requirements translate to: [[app(app(app(compose, _x0), _x1), _x2)]] = 24 + x2 + 2x1 + 4x0 > x2 + 2x0 + 2x1 = [[app(_x0, app(_x1, _x2))]] We can thus remove the following rules: app(app(app(compose, X), Y), Z) => app(X, app(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.
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