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SRS Standard pair #487089388
details
property
value
status
complete
benchmark
matchbox1.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n054.star.cs.uiowa.edu
space
Secret_05_SRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
13.2419 seconds
cpu usage
39.3395
user time
37.6989
system time
1.64056
max virtual memory
5.959758E7
max residence set size
4242092.0
stage attributes
key
value
starexec-result
YES
output
YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 33 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 12 ms] (4) QDP (5) MRRProof [EQUIVALENT, 123 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 242 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 192 ms] (10) QDP (11) DependencyGraphProof [EQUIVALENT, 0 ms] (12) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: r(e(x1)) -> w(r(x1)) i(t(x1)) -> e(r(x1)) e(w(x1)) -> r(i(x1)) t(e(x1)) -> r(e(x1)) w(r(x1)) -> i(t(x1)) e(r(x1)) -> e(w(x1)) r(i(t(e(r(x1))))) -> e(w(r(i(t(e(x1)))))) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(e(x_1)) = 1 + x_1 POL(i(x_1)) = x_1 POL(r(x_1)) = 2 + x_1 POL(t(x_1)) = 3 + x_1 POL(w(x_1)) = 1 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: t(e(x1)) -> r(e(x1)) e(r(x1)) -> e(w(x1)) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: r(e(x1)) -> w(r(x1)) i(t(x1)) -> e(r(x1)) e(w(x1)) -> r(i(x1)) w(r(x1)) -> i(t(x1)) r(i(t(e(r(x1))))) -> e(w(r(i(t(e(x1)))))) Q is empty. ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: R(e(x1)) -> W(r(x1)) R(e(x1)) -> R(x1) I(t(x1)) -> E(r(x1)) I(t(x1)) -> R(x1) E(w(x1)) -> R(i(x1)) E(w(x1)) -> I(x1) W(r(x1)) -> I(t(x1)) R(i(t(e(r(x1))))) -> E(w(r(i(t(e(x1)))))) R(i(t(e(r(x1))))) -> W(r(i(t(e(x1))))) R(i(t(e(r(x1))))) -> R(i(t(e(x1)))) R(i(t(e(r(x1))))) -> I(t(e(x1))) R(i(t(e(r(x1))))) -> E(x1) The TRS R consists of the following rules: r(e(x1)) -> w(r(x1))
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