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TRS Stand 20472 pair #381712877
details
property
value
status
complete
benchmark
rta3.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n013.star.cs.uiowa.edu
space
AProVE_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.66083097458 seconds
cpu usage
3.752901082
max memory
2.2523904E8
stage attributes
key
value
output-size
2652
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) QDPSizeChangeProof [EQUIVALENT, 0 ms] (4) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: ack(0, y) -> s(y) ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) f(s(x), y) -> f(x, s(x)) f(x, s(y)) -> f(y, x) f(x, y) -> ack(x, y) ack(s(x), y) -> f(x, x) 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: ACK(s(x), 0) -> ACK(x, s(0)) ACK(s(x), s(y)) -> ACK(x, ack(s(x), y)) ACK(s(x), s(y)) -> ACK(s(x), y) F(s(x), y) -> F(x, s(x)) F(x, s(y)) -> F(y, x) F(x, y) -> ACK(x, y) ACK(s(x), y) -> F(x, x) The TRS R consists of the following rules: ack(0, y) -> s(y) ack(s(x), 0) -> ack(x, s(0)) ack(s(x), s(y)) -> ack(x, ack(s(x), y)) f(s(x), y) -> f(x, s(x)) f(x, s(y)) -> f(y, x) f(x, y) -> ack(x, y) ack(s(x), y) -> f(x, x) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) 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. From the DPs we obtained the following set of size-change graphs: *ACK(s(x), s(y)) -> ACK(s(x), y) The graph contains the following edges 1 >= 1, 2 > 2 *F(x, y) -> ACK(x, y) The graph contains the following edges 1 >= 1, 2 >= 2 *ACK(s(x), s(y)) -> ACK(x, ack(s(x), y)) The graph contains the following edges 1 > 1 *ACK(s(x), y) -> F(x, x) The graph contains the following edges 1 > 1, 1 > 2 *ACK(s(x), 0) -> ACK(x, s(0)) The graph contains the following edges 1 > 1 *F(s(x), y) -> F(x, s(x)) The graph contains the following edges 1 > 1, 1 >= 2 *F(x, s(y)) -> F(y, x) The graph contains the following edges 2 > 1, 1 >= 2
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