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TRS Stand 20472 pair #381712939
details
property
value
status
complete
benchmark
lescanne.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n083.star.cs.uiowa.edu
space
Rubio_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0366361141205 seconds
cpu usage
0.03086102
max memory
1392640.0
stage attributes
key
value
output-size
2292
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. div : [o * o] --> o e : [] --> o i : [o] --> o div(X, e) => i(X) i(div(X, Y)) => div(Y, X) div(div(X, Y), Z) => div(Y, div(i(X), 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]): div(X, e) >? i(X) i(div(X, Y)) >? div(Y, X) div(div(X, Y), Z) >? div(Y, div(i(X), Z)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: div = \y0y1.y1 + 2y0 e = 3 i = \y0.2y0 Using this interpretation, the requirements translate to: [[div(_x0, e)]] = 3 + 2x0 > 2x0 = [[i(_x0)]] [[i(div(_x0, _x1))]] = 2x1 + 4x0 >= x0 + 2x1 = [[div(_x1, _x0)]] [[div(div(_x0, _x1), _x2)]] = x2 + 2x1 + 4x0 >= x2 + 2x1 + 4x0 = [[div(_x1, div(i(_x0), _x2))]] We can thus remove the following rules: div(X, e) => i(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]): i(div(X, Y)) >? div(Y, X) div(div(X, Y), Z) >? div(Y, div(i(X), Z)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: div = \y0y1.1 + y1 + 2y0 i = \y0.2y0 Using this interpretation, the requirements translate to: [[i(div(_x0, _x1))]] = 2 + 2x1 + 4x0 > 1 + x0 + 2x1 = [[div(_x1, _x0)]] [[div(div(_x0, _x1), _x2)]] = 3 + x2 + 2x1 + 4x0 > 2 + x2 + 2x1 + 4x0 = [[div(_x1, div(i(_x0), _x2))]] We can thus remove the following rules: i(div(X, Y)) => div(Y, X) div(div(X, Y), Z) => div(Y, div(i(X), 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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