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SRS Standard pair #487519841
details
property
value
status
complete
benchmark
z086.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n136.star.cs.uiowa.edu
space
Zantema_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
9.65936112404 seconds
cpu usage
23.20590457
max memory
3.393437696E9
stage attributes
key
value
output-size
4704
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 6 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 58 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 38 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(x1)) -> c(b(x1)) b(b(x1)) -> c(a(x1)) c(c(x1)) -> b(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(a(x1)) -> C(b(x1)) A(a(x1)) -> B(x1) B(b(x1)) -> C(a(x1)) B(b(x1)) -> A(x1) C(c(x1)) -> B(a(x1)) C(c(x1)) -> A(x1) The TRS R consists of the following rules: a(a(x1)) -> c(b(x1)) b(b(x1)) -> c(a(x1)) c(c(x1)) -> b(a(x1)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. A(a(x1)) -> C(b(x1)) A(a(x1)) -> B(x1) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( B_1(x_1) ) = max{0, x_1 - 2} POL( C_1(x_1) ) = max{0, x_1 - 2} POL( c_1(x_1) ) = 2x_1 + 2 POL( b_1(x_1) ) = 2x_1 + 2 POL( a_1(x_1) ) = 2x_1 + 2 POL( A_1(x_1) ) = x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: c(c(x1)) -> b(a(x1)) b(b(x1)) -> c(a(x1)) a(a(x1)) -> c(b(x1)) ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: B(b(x1)) -> C(a(x1))
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