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TRS Stand 20472 pair #381710641
details
property
value
status
complete
benchmark
ternary-hard.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n007.star.cs.uiowa.edu
space
CiME_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.97118806839 seconds
cpu usage
4.742547097
max memory
2.71540224E8
stage attributes
key
value
output-size
13843
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 28 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) UsableRulesProof [EQUIVALENT, 0 ms] (12) QDP (13) MRRProof [EQUIVALENT, 33 ms] (14) QDP (15) QDPSizeChangeProof [EQUIVALENT, 0 ms] (16) YES (17) QDP (18) QDPSizeChangeProof [EQUIVALENT, 0 ms] (19) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(#) -> # +(#, x) -> x +(x, #) -> x +(0(x), 0(y)) -> 0(+(x, y)) +(0(x), 1(y)) -> 1(+(x, y)) +(1(x), 0(y)) -> 1(+(x, y)) +(0(x), j(y)) -> j(+(x, y)) +(j(x), 0(y)) -> j(+(x, y)) +(1(x), 1(y)) -> j(+(+(x, y), 1(#))) +(j(x), j(y)) -> 1(+(+(x, y), j(#))) +(1(x), j(y)) -> 0(+(x, y)) +(j(x), 1(y)) -> 0(+(x, y)) +(+(x, y), z) -> +(x, +(y, z)) opp(#) -> # opp(0(x)) -> 0(opp(x)) opp(1(x)) -> j(opp(x)) opp(j(x)) -> 1(opp(x)) -(x, y) -> +(x, opp(y)) *(#, x) -> # *(0(x), y) -> 0(*(x, y)) *(1(x), y) -> +(0(*(x, y)), y) *(j(x), y) -> -(0(*(x, y)), y) *(*(x, y), z) -> *(x, *(y, z)) *(+(x, y), z) -> +(*(x, z), *(y, z)) *(x, +(y, z)) -> +(*(x, y), *(x, z)) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: +^1(0(x), 0(y)) -> 0^1(+(x, y)) +^1(0(x), 0(y)) -> +^1(x, y) +^1(0(x), 1(y)) -> +^1(x, y) +^1(1(x), 0(y)) -> +^1(x, y) +^1(0(x), j(y)) -> +^1(x, y) +^1(j(x), 0(y)) -> +^1(x, y) +^1(1(x), 1(y)) -> +^1(+(x, y), 1(#)) +^1(1(x), 1(y)) -> +^1(x, y) +^1(j(x), j(y)) -> +^1(+(x, y), j(#)) +^1(j(x), j(y)) -> +^1(x, y) +^1(1(x), j(y)) -> 0^1(+(x, y)) +^1(1(x), j(y)) -> +^1(x, y) +^1(j(x), 1(y)) -> 0^1(+(x, y)) +^1(j(x), 1(y)) -> +^1(x, y) +^1(+(x, y), z) -> +^1(x, +(y, z)) +^1(+(x, y), z) -> +^1(y, z) OPP(0(x)) -> 0^1(opp(x)) OPP(0(x)) -> OPP(x) OPP(1(x)) -> OPP(x)
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