details

property	value
status	complete
benchmark	sizeChange.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n186.star.cs.uiowa.edu
space	AProVE_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.74568 seconds
cpu usage	3.75262
user time	3.60236
system time	0.150265
max virtual memory	1.8343148E7
max residence set size	242948.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) UsableRulesProof [EQUIVALENT, 0 ms] (6) QDP (7) QReductionProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: r(xs, ys, zs, nil) -> xs r(xs, nil, zs, cons(w, ws)) -> r(xs, xs, cons(succ(zero), zs), ws) r(xs, cons(y, ys), nil, cons(w, ws)) -> r(xs, xs, cons(succ(zero), nil), ws) r(xs, cons(y, ys), cons(z, zs), cons(w, ws)) -> r(ys, cons(y, ys), zs, cons(succ(zero), cons(w, ws))) Q is empty. ---------------------------------------- (1) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: r(xs, ys, zs, nil) -> xs r(xs, nil, zs, cons(w, ws)) -> r(xs, xs, cons(succ(zero), zs), ws) r(xs, cons(y, ys), nil, cons(w, ws)) -> r(xs, xs, cons(succ(zero), nil), ws) r(xs, cons(y, ys), cons(z, zs), cons(w, ws)) -> r(ys, cons(y, ys), zs, cons(succ(zero), cons(w, ws))) The set Q consists of the following terms: r(x0, x1, x2, nil) r(x0, nil, x1, cons(x2, x3)) r(x0, cons(x1, x2), nil, cons(x3, x4)) r(x0, cons(x1, x2), cons(x3, x4), cons(x5, x6)) ---------------------------------------- (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(xs, nil, zs, cons(w, ws)) -> R(xs, xs, cons(succ(zero), zs), ws) R(xs, cons(y, ys), nil, cons(w, ws)) -> R(xs, xs, cons(succ(zero), nil), ws) R(xs, cons(y, ys), cons(z, zs), cons(w, ws)) -> R(ys, cons(y, ys), zs, cons(succ(zero), cons(w, ws))) The TRS R consists of the following rules: r(xs, ys, zs, nil) -> xs r(xs, nil, zs, cons(w, ws)) -> r(xs, xs, cons(succ(zero), zs), ws) r(xs, cons(y, ys), nil, cons(w, ws)) -> r(xs, xs, cons(succ(zero), nil), ws) r(xs, cons(y, ys), cons(z, zs), cons(w, ws)) -> r(ys, cons(y, ys), zs, cons(succ(zero), cons(w, ws))) The set Q consists of the following terms: r(x0, x1, x2, nil) r(x0, nil, x1, cons(x2, x3)) r(x0, cons(x1, x2), nil, cons(x3, x4)) r(x0, cons(x1, x2), cons(x3, x4), cons(x5, x6)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: R(xs, nil, zs, cons(w, ws)) -> R(xs, xs, cons(succ(zero), zs), ws) popout

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all output return to TRS Standard