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SRS Standard pair #516976702
details
property
value
status
complete
benchmark
random-376.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n071.star.cs.uiowa.edu
space
Waldmann_19
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
4.24456095695 seconds
cpu usage
13.558500485
max memory
1.240752128E9
stage attributes
key
value
output-size
4534
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 32 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 1 ms] (4) AND (5) QDP (6) QDPSizeChangeProof [EQUIVALENT, 0 ms] (7) YES (8) QDP (9) QDPOrderProof [EQUIVALENT, 60 ms] (10) QDP (11) PisEmptyProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(b(b(x1)))) -> b(a(a(b(x1)))) b(a(a(a(x1)))) -> b(b(a(a(x1)))) a(b(a(b(x1)))) -> b(b(a(a(x1)))) 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: A(b(b(b(x1)))) -> B(a(a(b(x1)))) A(b(b(b(x1)))) -> A(a(b(x1))) A(b(b(b(x1)))) -> A(b(x1)) B(a(a(a(x1)))) -> B(b(a(a(x1)))) B(a(a(a(x1)))) -> B(a(a(x1))) A(b(a(b(x1)))) -> B(b(a(a(x1)))) A(b(a(b(x1)))) -> B(a(a(x1))) A(b(a(b(x1)))) -> A(a(x1)) A(b(a(b(x1)))) -> A(x1) The TRS R consists of the following rules: a(b(b(b(x1)))) -> b(a(a(b(x1)))) b(a(a(a(x1)))) -> b(b(a(a(x1)))) a(b(a(b(x1)))) -> b(b(a(a(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: B(a(a(a(x1)))) -> B(a(a(x1))) The TRS R consists of the following rules: a(b(b(b(x1)))) -> b(a(a(b(x1)))) b(a(a(a(x1)))) -> b(b(a(a(x1)))) a(b(a(b(x1)))) -> b(b(a(a(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (6) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
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