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TRS Stand 20472 pair #381710146
details
property
value
status
complete
benchmark
cime1.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n073.star.cs.uiowa.edu
space
Secret_05_TRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
16.6316099167 seconds
cpu usage
38.095241129
max memory
3.671826432E9
stage attributes
key
value
output-size
22029
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 12 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 604 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) AND (9) QDP (10) QDPOrderProof [EQUIVALENT, 0 ms] (11) QDP (12) PisEmptyProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) QDPOrderProof [EQUIVALENT, 0 ms] (16) QDP (17) PisEmptyProof [EQUIVALENT, 0 ms] (18) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: sortSu(circ(sortSu(cons(te(a), sortSu(s))), sortSu(t))) -> sortSu(cons(te(msubst(te(a), sortSu(t))), sortSu(circ(sortSu(s), sortSu(t))))) sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(te(a), sortSu(t))))) -> sortSu(cons(te(a), sortSu(circ(sortSu(s), sortSu(t))))) sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(sop(lift), sortSu(t))))) -> sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t))))) sortSu(circ(sortSu(circ(sortSu(s), sortSu(t))), sortSu(u))) -> sortSu(circ(sortSu(s), sortSu(circ(sortSu(t), sortSu(u))))) sortSu(circ(sortSu(s), sortSu(id))) -> sortSu(s) sortSu(circ(sortSu(id), sortSu(s))) -> sortSu(s) sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(circ(sortSu(cons(sop(lift), sortSu(t))), sortSu(u))))) -> sortSu(circ(sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t))))), sortSu(u))) te(subst(te(a), sortSu(id))) -> te(a) te(msubst(te(a), sortSu(id))) -> te(a) te(msubst(te(msubst(te(a), sortSu(s))), sortSu(t))) -> te(msubst(te(a), sortSu(circ(sortSu(s), sortSu(t))))) 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: SORTSU(circ(sortSu(cons(te(a), sortSu(s))), sortSu(t))) -> SORTSU(cons(te(msubst(te(a), sortSu(t))), sortSu(circ(sortSu(s), sortSu(t))))) SORTSU(circ(sortSu(cons(te(a), sortSu(s))), sortSu(t))) -> TE(msubst(te(a), sortSu(t))) SORTSU(circ(sortSu(cons(te(a), sortSu(s))), sortSu(t))) -> SORTSU(circ(sortSu(s), sortSu(t))) SORTSU(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(te(a), sortSu(t))))) -> SORTSU(cons(te(a), sortSu(circ(sortSu(s), sortSu(t))))) SORTSU(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(te(a), sortSu(t))))) -> SORTSU(circ(sortSu(s), sortSu(t))) SORTSU(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(sop(lift), sortSu(t))))) -> SORTSU(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t))))) SORTSU(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(sop(lift), sortSu(t))))) -> SORTSU(circ(sortSu(s), sortSu(t))) SORTSU(circ(sortSu(circ(sortSu(s), sortSu(t))), sortSu(u))) -> SORTSU(circ(sortSu(s), sortSu(circ(sortSu(t), sortSu(u))))) SORTSU(circ(sortSu(circ(sortSu(s), sortSu(t))), sortSu(u))) -> SORTSU(circ(sortSu(t), sortSu(u))) SORTSU(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(circ(sortSu(cons(sop(lift), sortSu(t))), sortSu(u))))) -> SORTSU(circ(sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t))))), sortSu(u))) SORTSU(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(circ(sortSu(cons(sop(lift), sortSu(t))), sortSu(u))))) -> SORTSU(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t))))) SORTSU(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(circ(sortSu(cons(sop(lift), sortSu(t))), sortSu(u))))) -> SORTSU(circ(sortSu(s), sortSu(t))) TE(msubst(te(msubst(te(a), sortSu(s))), sortSu(t))) -> TE(msubst(te(a), sortSu(circ(sortSu(s), sortSu(t))))) TE(msubst(te(msubst(te(a), sortSu(s))), sortSu(t))) -> SORTSU(circ(sortSu(s), sortSu(t))) The TRS R consists of the following rules: sortSu(circ(sortSu(cons(te(a), sortSu(s))), sortSu(t))) -> sortSu(cons(te(msubst(te(a), sortSu(t))), sortSu(circ(sortSu(s), sortSu(t))))) sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(te(a), sortSu(t))))) -> sortSu(cons(te(a), sortSu(circ(sortSu(s), sortSu(t))))) sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(sop(lift), sortSu(t))))) -> sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t))))) sortSu(circ(sortSu(circ(sortSu(s), sortSu(t))), sortSu(u))) -> sortSu(circ(sortSu(s), sortSu(circ(sortSu(t), sortSu(u))))) sortSu(circ(sortSu(s), sortSu(id))) -> sortSu(s) sortSu(circ(sortSu(id), sortSu(s))) -> sortSu(s) sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(circ(sortSu(cons(sop(lift), sortSu(t))), sortSu(u))))) -> sortSu(circ(sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t))))), sortSu(u))) te(subst(te(a), sortSu(id))) -> te(a) te(msubst(te(a), sortSu(id))) -> te(a) te(msubst(te(msubst(te(a), sortSu(s))), sortSu(t))) -> te(msubst(te(a), sortSu(circ(sortSu(s), sortSu(t))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes. ----------------------------------------
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