Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
TRS Standard pair #487068818
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
n173.star.cs.uiowa.edu
space
AProVE_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.6899 seconds
cpu usage
3.74304
user time
3.5759
system time
0.167135
max virtual memory
1.8408912E7
max residence set size
239992.0
stage attributes
key
value
starexec-result
YES
output
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) QDPSizeChangeProof [EQUIVALENT,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 ---------------------------------------- (4) YES
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to TRS Standard