TRS Innermost pair #487092961

details

property	value
status	complete
benchmark	#4.30b.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n147.star.cs.uiowa.edu
space	AG01_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.06621 seconds
cpu usage	6.75459
user time	6.45978
system time	0.294813
max virtual memory	1.8677216E7
max residence set size	397656.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 28 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) Induction-Processor [SOUND, 6 ms]
        (27) AND
            (28) QDP
                (29) DependencyGraphProof [EQUIVALENT, 0 ms]
                (30) TRUE
            (31) QTRS
                (32) QTRSRRRProof [EQUIVALENT, 3 ms]
                (33) QTRS
                (34) RisEmptyProof [EQUIVALENT, 0 ms]
                (35) YES


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

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

   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   mod(0, y) -> 0
   mod(s(x), 0) -> 0
   mod(s(x), s(y)) -> if_mod(le(y, x), s(x), s(y))
   if_mod(true, x, y) -> mod(minus(x, y), y)
   if_mod(false, s(x), s(y)) -> s(x)

The set Q consists of the following terms:

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   minus(x0, 0)
   minus(s(x0), s(x1))
   mod(0, x0)
   mod(s(x0), 0)
   mod(s(x0), s(x1))
   if_mod(true, x0, x1)
   if_mod(false, s(x0), s(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:

   LE(s(x), s(y)) -> LE(x, y)
   MINUS(s(x), s(y)) -> MINUS(x, y)
   MOD(s(x), s(y)) -> IF_MOD(le(y, x), s(x), s(y))
   MOD(s(x), s(y)) -> LE(y, x)
   IF_MOD(true, x, y) -> MOD(minus(x, y), y)
   IF_MOD(true, x, y) -> MINUS(x, y)

The TRS R consists of the following rules:

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

popout

output may be truncated. 'popout' for the full output.

job log

popout

actions all output return to TRS Innermost