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TRS Stand 20472 pair #381712398
details
property
value
status
complete
benchmark
18.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n093.star.cs.uiowa.edu
space
Various_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0469260215759 seconds
cpu usage
0.031545722
max memory
1593344.0
stage attributes
key
value
output-size
2682
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 !times : [o * o] --> o u : [] --> o !plus(!times(X, Y), !times(X, Z)) => !times(X, !plus(Y, Z)) !plus(!plus(X, Y), Z) => !plus(X, !plus(Y, Z)) !plus(!times(X, Y), !plus(!times(X, Z), u)) => !plus(!times(X, !plus(Y, Z)), u) 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(!times(X, Y), !times(X, Z)) >? !times(X, !plus(Y, Z)) !plus(!plus(X, Y), Z) >? !plus(X, !plus(Y, Z)) !plus(!times(X, Y), !plus(!times(X, Z), u)) >? !plus(!times(X, !plus(Y, Z)), u) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !plus = \y0y1.y0 + y1 !times = \y0y1.1 + 2y0 + 2y1 u = 0 Using this interpretation, the requirements translate to: [[!plus(!times(_x0, _x1), !times(_x0, _x2))]] = 2 + 2x1 + 2x2 + 4x0 > 1 + 2x0 + 2x1 + 2x2 = [[!times(_x0, !plus(_x1, _x2))]] [[!plus(!plus(_x0, _x1), _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[!plus(_x0, !plus(_x1, _x2))]] [[!plus(!times(_x0, _x1), !plus(!times(_x0, _x2), u))]] = 2 + 2x1 + 2x2 + 4x0 > 1 + 2x0 + 2x1 + 2x2 = [[!plus(!times(_x0, !plus(_x1, _x2)), u)]] We can thus remove the following rules: !plus(!times(X, Y), !times(X, Z)) => !times(X, !plus(Y, Z)) !plus(!times(X, Y), !plus(!times(X, Z), u)) => !plus(!times(X, !plus(Y, Z)), u) 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.
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