TRS Innermost pair #487524001

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	n184.star.cs.uiowa.edu
space	AG01_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.18426990509 seconds
cpu usage	7.006600347
max memory	3.76188928E8

stage attributes

key	value
output-size	23395
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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


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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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))


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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(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)

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