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TRS Stand 20472 pair #381711565
details
property
value
status
complete
benchmark
jones4.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n039.star.cs.uiowa.edu
space
Mixed_TRS
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0235319137573 seconds
cpu usage
0.019143057
max memory
1302528.0
stage attributes
key
value
output-size
1574
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. 0 : [] --> o p : [o * o * o] --> o s : [o] --> o p(X, Y, s(Z)) => p(X, Z, Y) p(X, s(Y), 0) => p(0, Y, X) p(X, 0, 0) => 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]): p(X, Y, s(Z)) >? p(X, Z, Y) p(X, s(Y), 0) >? p(0, Y, X) p(X, 0, 0) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 p = \y0y1y2.3 + y2 + 2y0 + 3y1 s = \y0.3 + 3y0 Using this interpretation, the requirements translate to: [[p(_x0, _x1, s(_x2))]] = 6 + 2x0 + 3x1 + 3x2 > 3 + x1 + 2x0 + 3x2 = [[p(_x0, _x2, _x1)]] [[p(_x0, s(_x1), 0)]] = 12 + 2x0 + 9x1 > 3 + x0 + 3x1 = [[p(0, _x1, _x0)]] [[p(_x0, 0, 0)]] = 3 + 2x0 > x0 = [[_x0]] We can thus remove the following rules: p(X, Y, s(Z)) => p(X, Z, Y) p(X, s(Y), 0) => p(0, Y, X) p(X, 0, 0) => 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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