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SRS Stand 10685 pair #381719500
details
property
value
status
complete
benchmark
z080.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n101.star.cs.uiowa.edu
space
Zantema_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
8.66679811478 seconds
cpu usage
29.750024851
max memory
4.007632896E9
stage attributes
key
value
output-size
8095
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 2 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 660 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) AND (9) QDP (10) UsableRulesProof [EQUIVALENT, 0 ms] (11) QDP (12) QReductionProof [EQUIVALENT, 0 ms] (13) QDP (14) QDPSizeChangeProof [EQUIVALENT, 0 ms] (15) YES (16) QDP (17) UsableRulesProof [EQUIVALENT, 0 ms] (18) QDP (19) QReductionProof [EQUIVALENT, 0 ms] (20) QDP (21) QDPSizeChangeProof [EQUIVALENT, 0 ms] (22) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: A(b(x1)) -> b(a(B(A(x1)))) B(a(x1)) -> a(b(A(B(x1)))) A(a(x1)) -> x1 B(b(x1)) -> x1 Q is empty. ---------------------------------------- (1) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: A(b(x1)) -> b(a(B(A(x1)))) B(a(x1)) -> a(b(A(B(x1)))) A(a(x1)) -> x1 B(b(x1)) -> x1 The set Q consists of the following terms: A(b(x0)) B(a(x0)) A(a(x0)) B(b(x0)) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: A^1(b(x1)) -> B^1(A(x1)) A^1(b(x1)) -> A^1(x1) B^1(a(x1)) -> A^1(B(x1)) B^1(a(x1)) -> B^1(x1) The TRS R consists of the following rules: A(b(x1)) -> b(a(B(A(x1)))) B(a(x1)) -> a(b(A(B(x1)))) A(a(x1)) -> x1 B(b(x1)) -> x1 The set Q consists of the following terms:
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