SRS Standard pair #487091506

details

property	value
status	complete
benchmark	z125.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	Zantema_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.17569 seconds
cpu usage	9.3725
user time	8.87726
system time	0.495236
max virtual memory	4.0313832E7
max residence set size	1146716.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRS Reverse [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) QDPSizeChangeProof [EQUIVALENT, 1 ms]
        (11) YES
    (12) QDP
        (13) QDPOrderProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QDPOrderProof [EQUIVALENT, 14 ms]
        (18) QDP
        (19) PisEmptyProof [EQUIVALENT, 0 ms]
        (20) YES


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

   f(x1) -> n(c(n(a(x1))))
   c(f(x1)) -> f(n(a(c(x1))))
   n(a(x1)) -> c(x1)
   c(c(x1)) -> c(x1)
   n(s(x1)) -> f(s(s(x1)))
   n(f(x1)) -> f(n(x1))

Q is empty.

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(1) QTRS Reverse (EQUIVALENT)
We applied the QTRS Reverse Processor [REVERSE].
----------------------------------------

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

   f(x1) -> a(n(c(n(x1))))
   f(c(x1)) -> c(a(n(f(x1))))
   a(n(x1)) -> c(x1)
   c(c(x1)) -> c(x1)
   s(n(x1)) -> s(s(f(x1)))
   f(n(x1)) -> n(f(x1))

Q is empty.

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(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(x1) -> A(n(c(n(x1))))
   F(x1) -> C(n(x1))
   F(c(x1)) -> C(a(n(f(x1))))
   F(c(x1)) -> A(n(f(x1)))
   F(c(x1)) -> F(x1)
   A(n(x1)) -> C(x1)
   S(n(x1)) -> S(s(f(x1)))
   S(n(x1)) -> S(f(x1))
   S(n(x1)) -> F(x1)
   F(n(x1)) -> F(x1)

The TRS R consists of the following rules:

   f(x1) -> a(n(c(n(x1))))
   f(c(x1)) -> c(a(n(f(x1))))
   a(n(x1)) -> c(x1)
   c(c(x1)) -> c(x1)
   s(n(x1)) -> s(s(f(x1)))
   f(n(x1)) -> n(f(x1))

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