TRS Standard pair #487066923

details

property	value
status	complete
benchmark	#4.30a.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n185.star.cs.uiowa.edu
space	Strategy_removed_AG01

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.18508 seconds
cpu usage	4.9176
user time	4.70962
system time	0.207974
max virtual memory	1.8478552E7
max residence set size	339852.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) AAECC Innermost [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QReductionProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES
    (21) QDP
        (22) UsableRulesProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) QReductionProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) Induction-Processor [SOUND, 24 ms]
        (27) AND
            (28) QDP
                (29) PisEmptyProof [EQUIVALENT, 0 ms]
                (30) YES
            (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:

   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(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))

Q is empty.

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

(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))

The TRS R 2 is 
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)

The signature Sigma is {le_2, true, false}
----------------------------------------

(2)
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(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))

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(0, s(x0))

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