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TRS Stand 20472 pair #381712518
details
property
value
status
complete
benchmark
21.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n114.star.cs.uiowa.edu
space
Various_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.222321033478 seconds
cpu usage
0.219423766
max memory
4911104.0
stage attributes
key
value
output-size
13914
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. !plus : [o * o] --> o p1 : [] --> o p10 : [] --> o p2 : [] --> o p5 : [] --> o !plus(p1, p1) => p2 !plus(p1, !plus(p2, p2)) => p5 !plus(p5, p5) => p10 !plus(!plus(X, Y), Z) => !plus(X, !plus(Y, Z)) !plus(p1, !plus(p1, X)) => !plus(p2, X) !plus(p1, !plus(p2, !plus(p2, X))) => !plus(p5, X) !plus(p2, p1) => !plus(p1, p2) !plus(p2, !plus(p1, X)) => !plus(p1, !plus(p2, X)) !plus(p2, !plus(p2, p2)) => !plus(p1, p5) !plus(p2, !plus(p2, !plus(p2, X))) => !plus(p1, !plus(p5, X)) !plus(p5, p1) => !plus(p1, p5) !plus(p5, !plus(p1, X)) => !plus(p1, !plus(p5, X)) !plus(p5, p2) => !plus(p2, p5) !plus(p5, !plus(p2, X)) => !plus(p2, !plus(p5, X)) !plus(p5, !plus(p5, X)) => !plus(p10, X) !plus(p10, p1) => !plus(p1, p10) !plus(p10, !plus(p1, X)) => !plus(p1, !plus(p10, X)) !plus(p10, p2) => !plus(p2, p10) !plus(p10, !plus(p2, X)) => !plus(p2, !plus(p10, X)) !plus(p10, p5) => !plus(p5, p10) !plus(p10, !plus(p5, X)) => !plus(p5, !plus(p10, 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]): !plus(p1, p1) >? p2 !plus(p1, !plus(p2, p2)) >? p5 !plus(p5, p5) >? p10 !plus(!plus(X, Y), Z) >? !plus(X, !plus(Y, Z)) !plus(p1, !plus(p1, X)) >? !plus(p2, X) !plus(p1, !plus(p2, !plus(p2, X))) >? !plus(p5, X) !plus(p2, p1) >? !plus(p1, p2) !plus(p2, !plus(p1, X)) >? !plus(p1, !plus(p2, X)) !plus(p2, !plus(p2, p2)) >? !plus(p1, p5) !plus(p2, !plus(p2, !plus(p2, X))) >? !plus(p1, !plus(p5, X)) !plus(p5, p1) >? !plus(p1, p5) !plus(p5, !plus(p1, X)) >? !plus(p1, !plus(p5, X)) !plus(p5, p2) >? !plus(p2, p5) !plus(p5, !plus(p2, X)) >? !plus(p2, !plus(p5, X)) !plus(p5, !plus(p5, X)) >? !plus(p10, X) !plus(p10, p1) >? !plus(p1, p10) !plus(p10, !plus(p1, X)) >? !plus(p1, !plus(p10, X)) !plus(p10, p2) >? !plus(p2, p10) !plus(p10, !plus(p2, X)) >? !plus(p2, !plus(p10, X)) !plus(p10, p5) >? !plus(p5, p10) !plus(p10, !plus(p5, X)) >? !plus(p5, !plus(p10, X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !plus = \y0y1.2 + y0 + y1 p1 = 0 p10 = 0 p2 = 0 p5 = 0 Using this interpretation, the requirements translate to: [[!plus(p1, p1)]] = 2 > 0 = [[p2]] [[!plus(p1, !plus(p2, p2))]] = 4 > 0 = [[p5]] [[!plus(p5, p5)]] = 2 > 0 = [[p10]] [[!plus(!plus(_x0, _x1), _x2)]] = 4 + x0 + x1 + x2 >= 4 + x0 + x1 + x2 = [[!plus(_x0, !plus(_x1, _x2))]] [[!plus(p1, !plus(p1, _x0))]] = 4 + x0 > 2 + x0 = [[!plus(p2, _x0)]] [[!plus(p1, !plus(p2, !plus(p2, _x0)))]] = 6 + x0 > 2 + x0 = [[!plus(p5, _x0)]] [[!plus(p2, p1)]] = 2 >= 2 = [[!plus(p1, p2)]] [[!plus(p2, !plus(p1, _x0))]] = 4 + x0 >= 4 + x0 = [[!plus(p1, !plus(p2, _x0))]] [[!plus(p2, !plus(p2, p2))]] = 4 > 2 = [[!plus(p1, p5)]] [[!plus(p2, !plus(p2, !plus(p2, _x0)))]] = 6 + x0 > 4 + x0 = [[!plus(p1, !plus(p5, _x0))]] [[!plus(p5, p1)]] = 2 >= 2 = [[!plus(p1, p5)]] [[!plus(p5, !plus(p1, _x0))]] = 4 + x0 >= 4 + x0 = [[!plus(p1, !plus(p5, _x0))]] [[!plus(p5, p2)]] = 2 >= 2 = [[!plus(p2, p5)]] [[!plus(p5, !plus(p2, _x0))]] = 4 + x0 >= 4 + x0 = [[!plus(p2, !plus(p5, _x0))]] [[!plus(p5, !plus(p5, _x0))]] = 4 + x0 > 2 + x0 = [[!plus(p10, _x0)]] [[!plus(p10, p1)]] = 2 >= 2 = [[!plus(p1, p10)]] [[!plus(p10, !plus(p1, _x0))]] = 4 + x0 >= 4 + x0 = [[!plus(p1, !plus(p10, _x0))]] [[!plus(p10, p2)]] = 2 >= 2 = [[!plus(p2, p10)]] [[!plus(p10, !plus(p2, _x0))]] = 4 + x0 >= 4 + x0 = [[!plus(p2, !plus(p10, _x0))]] [[!plus(p10, p5)]] = 2 >= 2 = [[!plus(p5, p10)]] [[!plus(p10, !plus(p5, _x0))]] = 4 + x0 >= 4 + x0 = [[!plus(p5, !plus(p10, _x0))]] We can thus remove the following rules:
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