TRS Standard pair #487068808

details

property	value
status	complete
benchmark	IJCAR_18.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n187.star.cs.uiowa.edu
space	AProVE_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.31495 seconds
cpu usage	5.02423
user time	4.77866
system time	0.24557
max virtual memory	5.4803068E7
max residence set size	348772.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 1 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) UsableRulesProof [EQUIVALENT, 0 ms]
        (12) QDP
        (13) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (14) YES
    (15) QDP
        (16) UsableRulesProof [EQUIVALENT, 0 ms]
        (17) QDP
        (18) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (19) YES
    (20) QDP
        (21) QDPOrderProof [EQUIVALENT, 14 ms]
        (22) QDP
        (23) QDPOrderProof [EQUIVALENT, 0 ms]
        (24) QDP
        (25) TransformationProof [EQUIVALENT, 0 ms]
        (26) QDP
        (27) DependencyGraphProof [EQUIVALENT, 0 ms]
        (28) TRUE
    (29) QDP
        (30) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (31) YES


----------------------------------------

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   plus(x, 0) -> x
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   times(0, y) -> 0
   times(s(0), y) -> y
   times(s(x), y) -> plus(y, times(x, y))
   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)))
   div(div(x, y), z) -> div(x, times(y, z))
   eq(0, 0) -> true
   eq(s(x), 0) -> false
   eq(0, s(y)) -> false
   eq(s(x), s(y)) -> eq(x, y)
   divides(y, x) -> eq(x, times(div(x, y), y))
   prime(s(s(x))) -> pr(s(s(x)), s(x))
   pr(x, s(0)) -> true
   pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
   if(true, x, y) -> false
   if(false, x, y) -> pr(x, y)

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:

   PLUS(s(x), y) -> PLUS(x, y)
   TIMES(s(x), y) -> PLUS(y, times(x, y))
   TIMES(s(x), y) -> TIMES(x, y)
   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))
   DIV(div(x, y), z) -> DIV(x, times(y, z))
   DIV(div(x, y), z) -> TIMES(y, z)
   EQ(s(x), s(y)) -> EQ(x, y)
   DIVIDES(y, x) -> EQ(x, times(div(x, y), y))
   DIVIDES(y, x) -> TIMES(div(x, y), y)
   DIVIDES(y, x) -> DIV(x, y)
   PRIME(s(s(x))) -> PR(s(s(x)), s(x))
   PR(x, s(s(y))) -> IF(divides(s(s(y)), x), x, s(y))
   PR(x, s(s(y))) -> DIVIDES(s(s(y)), x)
   IF(false, x, y) -> PR(x, y)

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