details

property	value
status	complete
benchmark	Applicative_first_order_05__#3.40.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n049.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	sol 37957
configuration	default
runtime (wallclock)	12.8597519398 seconds
cpu usage	16.559880763
max memory	2.38096384E8

stage attributes

key	value
output-size	30325
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES We split firstr-order part and higher-order part, and do modular checking by a general modularity. ******** FO SN check ******** Check SN using AProVE (RWTH Aachen University) proof of tmp15483.trs # AProVE Commit ID: 240871ee8d33536d563834eff18151406a8bc3fe ffrohn 20170821 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0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) MRRProof [EQUIVALENT, 15 ms] (9) QDP (10) MRRProof [EQUIVALENT, 24 ms] (11) QDP (12) MRRProof [EQUIVALENT, 0 ms] (13) QDP (14) DependencyGraphProof [EQUIVALENT, 0 ms] (15) TRUE (16) QDP (17) UsableRulesProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPSizeChangeProof [EQUIVALENT, 0 ms] (20) YES (21) QDP (22) QDPOrderProof [EQUIVALENT, 0 ms] (23) QDP (24) PisEmptyProof [EQUIVALENT, 0 ms] (25) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: minus(X, 0) -> X minus(s(Y), s(U)) -> minus(Y, U) quot(0, s(V)) -> 0 quot(s(W), s(P)) -> s(quot(minus(W, P), s(P))) plus(0, X1) -> X1 plus(s(Y1), U1) -> s(plus(Y1, U1)) plus(minus(V1, s(0)), minus(W1, s(s(P1)))) -> plus(minus(W1, s(s(P1))), minus(V1, s(0))) plus(plus(X2, s(0)), plus(Y2, s(s(U2)))) -> plus(plus(Y2, s(s(U2))), plus(X2, s(0))) _(X1, X2) -> X1 _(X1, X2) -> X2 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: MINUS(s(Y), s(U)) -> MINUS(Y, U) QUOT(s(W), s(P)) -> QUOT(minus(W, P), s(P)) QUOT(s(W), s(P)) -> MINUS(W, P) PLUS(s(Y1), U1) -> PLUS(Y1, U1) PLUS(minus(V1, s(0)), minus(W1, s(s(P1)))) -> PLUS(minus(W1, s(s(P1))), minus(V1, s(0))) PLUS(plus(X2, s(0)), plus(Y2, s(s(U2)))) -> PLUS(plus(Y2, s(s(U2))), plus(X2, s(0))) The TRS R consists of the following rules: minus(X, 0) -> X minus(s(Y), s(U)) -> minus(Y, U) quot(0, s(V)) -> 0 quot(s(W), s(P)) -> s(quot(minus(W, P), s(P))) plus(0, X1) -> X1 plus(s(Y1), U1) -> s(plus(Y1, U1)) plus(minus(V1, s(0)), minus(W1, s(s(P1)))) -> plus(minus(W1, s(s(P1))), minus(V1, s(0))) plus(plus(X2, s(0)), plus(Y2, s(s(U2)))) -> plus(plus(Y2, s(s(U2))), plus(X2, s(0))) _(X1, X2) -> X1 _(X1, X2) -> X2 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- popout

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job log

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actions

all output return to HRS union beta 16688