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	n135.star.cs.uiowa.edu
space	AProVE_06

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.18658304214 seconds
cpu usage	5.614145988
max memory	4.71359488E8

stage attributes

key	value
output-size	5488
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof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, 14 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. ---------------------------------------- popout

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actions

all output return to TRS Standard