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TRS Equat 89423 pair #381732773
details
property
value
status
complete
benchmark
bag-sum-prod-distr.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n052.star.cs.uiowa.edu
space
Mixed_AC
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.83879208565 seconds
cpu usage
4.069354376
max memory
2.46439936E8
stage attributes
key
value
output-size
7941
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given ETRS could be proven: (0) ETRS (1) RRRPoloETRSProof [EQUIVALENT, 153 ms] (2) ETRS (3) RRRPoloETRSProof [EQUIVALENT, 0 ms] (4) ETRS (5) RRRPoloETRSProof [EQUIVALENT, 9 ms] (6) ETRS (7) RRRPoloETRSProof [EQUIVALENT, 0 ms] (8) ETRS (9) RisEmptyProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Equational rewrite system: The TRS R consists of the following rules: 0(#) -> # +(#, x) -> x +(0(x), 0(y)) -> 0(+(x, y)) +(0(x), 1(y)) -> 1(+(x, y)) +(1(x), 1(y)) -> 0(+(1(#), +(x, y))) *(#, x) -> # *(0(x), y) -> 0(*(x, y)) *(1(x), y) -> +(0(*(x, y)), y) *(+(y, z), x) -> +(*(x, y), *(x, z)) U(empty, b) -> b sum(empty) -> 0(#) sum(singl(x)) -> x sum(U(x, y)) -> +(sum(x), sum(y)) prod(empty) -> 1(#) prod(singl(x)) -> x prod(U(x, y)) -> *(prod(x), prod(y)) The set E consists of the following equations: *(x, y) == *(y, x) +(x, y) == +(y, x) U(x, y) == U(y, x) *(*(x, y), z) == *(x, *(y, z)) +(+(x, y), z) == +(x, +(y, z)) U(U(x, y), z) == U(x, U(y, z)) ---------------------------------------- (1) RRRPoloETRSProof (EQUIVALENT) The following E TRS is given: Equational rewrite system: The TRS R consists of the following rules: 0(#) -> # +(#, x) -> x +(0(x), 0(y)) -> 0(+(x, y)) +(0(x), 1(y)) -> 1(+(x, y)) +(1(x), 1(y)) -> 0(+(1(#), +(x, y))) *(#, x) -> # *(0(x), y) -> 0(*(x, y)) *(1(x), y) -> +(0(*(x, y)), y) *(+(y, z), x) -> +(*(x, y), *(x, z)) U(empty, b) -> b sum(empty) -> 0(#) sum(singl(x)) -> x sum(U(x, y)) -> +(sum(x), sum(y)) prod(empty) -> 1(#) prod(singl(x)) -> x prod(U(x, y)) -> *(prod(x), prod(y)) The set E consists of the following equations: *(x, y) == *(y, x) +(x, y) == +(y, x) U(x, y) == U(y, x) *(*(x, y), z) == *(x, *(y, z)) +(+(x, y), z) == +(x, +(y, z)) U(U(x, y), z) == U(x, U(y, z)) The following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly by a polynomial ordering: +(#, x) -> x *(#, x) -> # *(1(x), y) -> +(0(*(x, y)), y) U(empty, b) -> b sum(empty) -> 0(#) sum(singl(x)) -> x
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