details

property	value
status	complete
benchmark	IJCAR_1.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n192.star.cs.uiowa.edu
space	AProVE_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.91545 seconds
cpu usage	3.81516
user time	3.64329
system time	0.171868
max virtual memory	1.8408912E7
max residence set size	246436.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 4 ms] (4) QDP (5) TransformationProof [EQUIVALENT, 0 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: div(0, y) -> 0 div(x, y) -> quot(x, y, y) quot(0, s(y), z) -> 0 quot(s(x), s(y), z) -> quot(x, y, z) quot(x, 0, s(z)) -> s(div(x, s(z))) 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: DIV(x, y) -> QUOT(x, y, y) QUOT(s(x), s(y), z) -> QUOT(x, y, z) QUOT(x, 0, s(z)) -> DIV(x, s(z)) The TRS R consists of the following rules: div(0, y) -> 0 div(x, y) -> quot(x, y, y) quot(0, s(y), z) -> 0 quot(s(x), s(y), z) -> quot(x, y, z) quot(x, 0, s(z)) -> s(div(x, s(z))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. QUOT(s(x), s(y), z) -> QUOT(x, y, z) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( DIV_2(x_1, x_2) ) = x_1 POL( QUOT_3(x_1, ..., x_3) ) = x_1 POL( s_1(x_1) ) = x_1 + 1 POL( 0 ) = 2 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: none ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: DIV(x, y) -> QUOT(x, y, y) QUOT(x, 0, s(z)) -> DIV(x, s(z)) The TRS R consists of the following rules: div(0, y) -> 0 div(x, y) -> quot(x, y, y) quot(0, s(y), z) -> 0 quot(s(x), s(y), z) -> quot(x, y, z) quot(x, 0, s(z)) -> s(div(x, s(z))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- popout

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