TRS Standard pair #487068308

details

property	value
status	complete
benchmark	t013.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n190.star.cs.uiowa.edu
space	HirokawaMiddeldorp_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.03893 seconds
cpu usage	4.28713
user time	4.10008
system time	0.187052
max virtual memory	1.8544088E7
max residence set size	261604.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 12 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) QDPOrderProof [EQUIVALENT, 34 ms]
        (16) QDP
        (17) PisEmptyProof [EQUIVALENT, 0 ms]
        (18) YES


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

   -(x, 0) -> x
   -(0, s(y)) -> 0
   -(s(x), s(y)) -> -(x, y)
   f(0) -> 0
   f(s(x)) -> -(s(x), g(f(x)))
   g(0) -> s(0)
   g(s(x)) -> -(s(x), f(g(x)))

Q is empty.

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

(1) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   -(x, 0) -> x
   -(0, s(y)) -> 0
   -(s(x), s(y)) -> -(x, y)
   f(0) -> 0
   f(s(x)) -> -(s(x), g(f(x)))
   g(0) -> s(0)
   g(s(x)) -> -(s(x), f(g(x)))

The set Q consists of the following terms:

   -(x0, 0)
   -(0, s(x0))
   -(s(x0), s(x1))
   f(0)
   f(s(x0))
   g(0)
   g(s(x0))


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

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

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   -^1(s(x), s(y)) -> -^1(x, y)
   F(s(x)) -> -^1(s(x), g(f(x)))
   F(s(x)) -> G(f(x))
   F(s(x)) -> F(x)
   G(s(x)) -> -^1(s(x), f(g(x)))
   G(s(x)) -> F(g(x))
   G(s(x)) -> G(x)

The TRS R consists of the following rules:

   -(x, 0) -> x
   -(0, s(y)) -> 0
   -(s(x), s(y)) -> -(x, y)
   f(0) -> 0

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