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TRS Standard pair #516965725
details
property
value
status
complete
benchmark
z09.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n093.star.cs.uiowa.edu
space
Zantema_05
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
6.06768393517 seconds
cpu usage
9.635210805
max memory
6.23091712E8
stage attributes
key
value
output-size
4846
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 746 ms] (6) QDP (7) QDPSizeChangeProof [EQUIVALENT, 0 ms] (8) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(lambda(x), y) -> lambda(a(x, p(1, a(y, t)))) a(p(x, y), z) -> p(a(x, z), a(y, z)) a(a(x, y), z) -> a(x, a(y, z)) lambda(x) -> x a(x, y) -> x a(x, y) -> y p(x, y) -> x p(x, y) -> y 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(lambda(x), y) -> LAMBDA(a(x, p(1, a(y, t)))) A(lambda(x), y) -> A(x, p(1, a(y, t))) A(lambda(x), y) -> P(1, a(y, t)) A(lambda(x), y) -> A(y, t) A(p(x, y), z) -> P(a(x, z), a(y, z)) A(p(x, y), z) -> A(x, z) A(p(x, y), z) -> A(y, z) A(a(x, y), z) -> A(x, a(y, z)) A(a(x, y), z) -> A(y, z) The TRS R consists of the following rules: a(lambda(x), y) -> lambda(a(x, p(1, a(y, t)))) a(p(x, y), z) -> p(a(x, z), a(y, z)) a(a(x, y), z) -> a(x, a(y, z)) lambda(x) -> x a(x, y) -> x a(x, y) -> y p(x, y) -> x p(x, y) -> y 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 1 SCC with 3 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: A(lambda(x), y) -> A(y, t) A(lambda(x), y) -> A(x, p(1, a(y, t))) A(p(x, y), z) -> A(x, z) A(p(x, y), z) -> A(y, z) A(a(x, y), z) -> A(x, a(y, z)) A(a(x, y), z) -> A(y, z) The TRS R consists of the following rules: a(lambda(x), y) -> lambda(a(x, p(1, a(y, t)))) a(p(x, y), z) -> p(a(x, z), a(y, z)) a(a(x, y), z) -> a(x, a(y, z)) lambda(x) -> x
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