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	n003.star.cs.uiowa.edu
space	Secret_05_TRS

run statistics

property	value
solver	AProVE23
configuration	standard
runtime (wallclock)	16.4413950443 seconds
cpu usage	41.426598093
max memory	1.031172096E9

stage attributes

key	value
output-size	22030
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# AProVE Commit ID: 363435ff5b19173ec8a83b45cfdec073678dabcc jan-christoph 20230815 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 36 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 676 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. ---------------------------------------- popout

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