TRS Innermost pair #487092943

details

property	value
status	complete
benchmark	#4.30.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n140.star.cs.uiowa.edu
space	AG01_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.32467 seconds
cpu usage	8.19502
user time	7.92717
system time	0.267851
max virtual memory	1.8549188E7
max residence set size	401560.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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


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

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

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   quot(x, s(y)) -> if_quot(le(s(y), x), x, s(y))
   if_quot(true, x, y) -> s(quot(minus(x, y), y))
   if_quot(false, x, y) -> 0

The set Q consists of the following terms:

   minus(x0, 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   quot(x0, s(x1))
   if_quot(true, x0, x1)
   if_quot(false, x0, x1)


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

(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(x), s(y)) -> MINUS(x, y)
   LE(s(x), s(y)) -> LE(x, y)
   QUOT(x, s(y)) -> IF_QUOT(le(s(y), x), x, s(y))
   QUOT(x, s(y)) -> LE(s(y), x)
   IF_QUOT(true, x, y) -> QUOT(minus(x, y), y)
   IF_QUOT(true, x, y) -> MINUS(x, y)

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   quot(x, s(y)) -> if_quot(le(s(y), x), x, s(y))
   if_quot(true, x, y) -> s(quot(minus(x, y), y))

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