TRS Standard pair #487066948

details

property	value
status	complete
benchmark	#4.32.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n072.star.cs.uiowa.edu
space	Strategy_removed_AG01

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.64491 seconds
cpu usage	3.56482
user time	3.4163
system time	0.148522
max virtual memory	1.8408912E7
max residence set size	236856.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) 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) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QReductionProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES


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

   f(x, c(y)) -> f(x, s(f(y, y)))
   f(s(x), s(y)) -> f(x, s(c(s(y))))

Q is empty.

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(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:

   f(x, c(y)) -> f(x, s(f(y, y)))
   f(s(x), s(y)) -> f(x, s(c(s(y))))

The set Q consists of the following terms:

   f(x0, c(x1))
   f(s(x0), s(x1))


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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(x, c(y)) -> F(x, s(f(y, y)))
   F(x, c(y)) -> F(y, y)
   F(s(x), s(y)) -> F(x, s(c(s(y))))

The TRS R consists of the following rules:

   f(x, c(y)) -> f(x, s(f(y, y)))
   f(s(x), s(y)) -> f(x, s(c(s(y))))

The set Q consists of the following terms:

   f(x0, c(x1))
   f(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
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(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
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(6)
Complex Obligation (AND)

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(7)

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