TRS Standard pair #615304716

details

property	value
status	complete
benchmark	#3.4.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n031.star.cs.uiowa.edu
space	AG01

run statistics

property	value
solver	AProVE23
configuration	standard
runtime (wallclock)	1.83640885353 seconds
cpu usage	4.655853404
max memory	2.6847232E8

stage attributes

key	value
output-size	8213
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# AProVE Commit ID: 363435ff5b19173ec8a83b45cfdec073678dabcc jan-christoph 20230815 unpublished


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) UsableRulesProof [EQUIVALENT, 0 ms]
        (12) QDP
        (13) MNOCProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (16) YES
    (17) QDP
        (18) QDPOrderProof [EQUIVALENT, 13 ms]
        (19) QDP
        (20) PisEmptyProof [EQUIVALENT, 0 ms]
        (21) 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)
   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   minus(minus(x, y), z) -> minus(x, plus(y, z))

Q is empty.

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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:

   MINUS(s(x), s(y)) -> MINUS(x, y)
   QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
   QUOT(s(x), s(y)) -> MINUS(x, y)
   PLUS(s(x), y) -> PLUS(x, y)
   MINUS(minus(x, y), z) -> MINUS(x, plus(y, z))
   MINUS(minus(x, y), z) -> PLUS(y, z)

The TRS R consists of the following rules:

   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(x, y), s(y)))
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   minus(minus(x, y), z) -> minus(x, plus(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 2 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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