TRS Standard pair #516967265

details

property	value
status	complete
benchmark	#3.36.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n088.star.cs.uiowa.edu
space	AG01

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.40431880951 seconds
cpu usage	6.475646734
max memory	4.80825344E8

stage attributes

key	value
output-size	7180
starexec-result	YES

output

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


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


YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished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, 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) QDPOrderProof [EQUIVALENT, 58 ms]
        (16) QDP
        (17) DependencyGraphProof [EQUIVALENT, 0 ms]
        (18) TRUE


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

(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)
   f(0) -> s(0)
   f(s(x)) -> minus(s(x), g(f(x)))
   g(0) -> 0
   g(s(x)) -> minus(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:

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

The set Q consists of the following terms:

   minus(x0, 0)
   minus(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:

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

The TRS R consists of the following rules:

   minus(x, 0) -> x

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