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TRS Standard pair #516962180
details
property
value
status
complete
benchmark
Ex23_Luc06_C.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n135.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
2.2514090538 seconds
cpu usage
5.805228796
max memory
4.87227392E8
stage attributes
key
value
output-size
7284
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) RFCMatchBoundsTRSProof [EQUIVALENT, 40 ms] (4) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: active(f(f(a))) -> mark(c(f(g(f(a))))) active(f(X)) -> f(active(X)) active(g(X)) -> g(active(X)) f(mark(X)) -> mark(f(X)) g(mark(X)) -> mark(g(X)) proper(f(X)) -> f(proper(X)) proper(a) -> ok(a) proper(c(X)) -> c(proper(X)) proper(g(X)) -> g(proper(X)) f(ok(X)) -> ok(f(X)) c(ok(X)) -> ok(c(X)) g(ok(X)) -> ok(g(X)) top(mark(X)) -> top(proper(X)) top(ok(X)) -> top(active(X)) Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a'(f(f(active(x)))) -> a'(f(g(f(c(mark(x)))))) f(active(X)) -> active(f(X)) g(active(X)) -> active(g(X)) mark(f(X)) -> f(mark(X)) mark(g(X)) -> g(mark(X)) f(proper(X)) -> proper(f(X)) a'(proper(x)) -> a'(ok(x)) c(proper(X)) -> proper(c(X)) g(proper(X)) -> proper(g(X)) ok(f(X)) -> f(ok(X)) ok(c(X)) -> c(ok(X)) ok(g(X)) -> g(ok(X)) mark(top(X)) -> proper(top(X)) ok(top(X)) -> active(top(X)) Q is empty. ---------------------------------------- (3) RFCMatchBoundsTRSProof (EQUIVALENT) Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 6. This implies Q-termination of R. The following rules were used to construct the certificate: a'(f(f(active(x)))) -> a'(f(g(f(c(mark(x)))))) f(active(X)) -> active(f(X)) g(active(X)) -> active(g(X)) mark(f(X)) -> f(mark(X)) mark(g(X)) -> g(mark(X)) f(proper(X)) -> proper(f(X)) a'(proper(x)) -> a'(ok(x)) c(proper(X)) -> proper(c(X)) g(proper(X)) -> proper(g(X)) ok(f(X)) -> f(ok(X)) ok(c(X)) -> c(ok(X)) ok(g(X)) -> g(ok(X)) mark(top(X)) -> proper(top(X)) ok(top(X)) -> active(top(X)) The certificate found is represented by the following graph. The certificate consists of the following enumerated nodes: 18, 19, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 67, 68, 72, 73, 74, 78, 79, 80, 81, 82, 87, 88, 89, 90, 91, 92, 93, 94, 95, 97, 98, 99 Node 18 is start node and node 19 is final node. Those nodes are connected through the following edges:
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