TRS Standard pair #615306634

details

property	value
status	complete
benchmark	#4.22.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n023.star.cs.uiowa.edu
space	Strategy_removed_AG01

run statistics

property	value
solver	AProVE23
configuration	standard
runtime (wallclock)	1.6796541214 seconds
cpu usage	4.160474336
max memory	2.6073088E8

stage attributes

key	value
output-size	3949
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# AProVE Commit ID: 363435ff5b19173ec8a83b45cfdec073678dabcc jan-christoph 20230815 unpublished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, 14 ms]
(4) QDP
(5) UsableRulesProof [EQUIVALENT, 0 ms]
(6) QDP
(7) QReductionProof [EQUIVALENT, 0 ms]
(8) QDP
(9) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(10) YES


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

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

   quot(0, s(y), s(z)) -> 0
   quot(s(x), s(y), z) -> quot(x, y, z)
   quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))

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:

   quot(0, s(y), s(z)) -> 0
   quot(s(x), s(y), z) -> quot(x, y, z)
   quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))

The set Q consists of the following terms:

   quot(0, s(x0), s(x1))
   quot(s(x0), s(x1), x2)
   quot(x0, 0, s(x1))


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

(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:

   QUOT(s(x), s(y), z) -> QUOT(x, y, z)
   QUOT(x, 0, s(z)) -> QUOT(x, s(z), s(z))

The TRS R consists of the following rules:

   quot(0, s(y), s(z)) -> 0
   quot(s(x), s(y), z) -> quot(x, y, z)
   quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))

The set Q consists of the following terms:

   quot(0, s(x0), s(x1))
   quot(s(x0), s(x1), x2)
   quot(x0, 0, s(x1))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(5) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   QUOT(s(x), s(y), z) -> QUOT(x, y, z)

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