TRS Standard pair #516968100

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	n177.star.cs.uiowa.edu
space	HirokawaMiddeldorp_04

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.34808802605 seconds
cpu usage	6.161819696
max memory	4.71977984E8

stage attributes

key	value
output-size	7121
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) Overlay + Local Confluence [EQUIVALENT, 14 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) QDPOrderProof [EQUIVALENT, 71 ms]
        (16) QDP
        (17) PisEmptyProof [EQUIVALENT, 0 ms]
        (18) YES


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

(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)

popout

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

job log

popout

actions all output return to TRS Standard