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TRS Inner 89993 pair #381733510
details
property
value
status
complete
benchmark
#4.36.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n036.star.cs.uiowa.edu
space
AG01_innermost
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.18608403206 seconds
cpu usage
5.476541201
max memory
4.02620416E8
stage attributes
key
value
output-size
30269
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QReductionProof [EQUIVALENT, 1 ms] (9) QDP (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) QReductionProof [EQUIVALENT, 2 ms] (16) QDP (17) QDPOrderProof [EQUIVALENT, 27 ms] (18) QDP (19) DependencyGraphProof [EQUIVALENT, 0 ms] (20) TRUE (21) QDP (22) UsableRulesProof [EQUIVALENT, 0 ms] (23) QDP (24) QReductionProof [EQUIVALENT, 1 ms] (25) QDP (26) QDPSizeChangeProof [EQUIVALENT, 0 ms] (27) YES (28) QDP (29) UsableRulesProof [EQUIVALENT, 0 ms] (30) QDP (31) QReductionProof [EQUIVALENT, 0 ms] (32) QDP (33) QDPSizeChangeProof [EQUIVALENT, 0 ms] (34) YES (35) QDP (36) UsableRulesProof [EQUIVALENT, 0 ms] (37) QDP (38) QReductionProof [EQUIVALENT, 1 ms] (39) QDP (40) QDPOrderProof [EQUIVALENT, 42 ms] (41) QDP (42) PisEmptyProof [EQUIVALENT, 0 ms] (43) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: eq(0, 0) -> true eq(0, s(m)) -> false eq(s(n), 0) -> false eq(s(n), s(m)) -> eq(n, m) le(0, m) -> true le(s(n), 0) -> false le(s(n), s(m)) -> le(n, m) min(cons(0, nil)) -> 0 min(cons(s(n), nil)) -> s(n) min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x))) if_min(true, cons(n, cons(m, x))) -> min(cons(n, x)) if_min(false, cons(n, cons(m, x))) -> min(cons(m, x)) replace(n, m, nil) -> nil replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x)) if_replace(true, n, m, cons(k, x)) -> cons(m, x) if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x)) sort(nil) -> nil sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x))) The set Q consists of the following terms: eq(0, 0) eq(0, s(x0)) eq(s(x0), 0) eq(s(x0), s(x1)) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) min(cons(0, nil)) min(cons(s(x0), nil)) min(cons(x0, cons(x1, x2))) if_min(true, cons(x0, cons(x1, x2))) if_min(false, cons(x0, cons(x1, x2))) replace(x0, x1, nil)
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