TRS Inner 89993 pair #381733449

details

property	value
status	complete
benchmark	#4.19.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n087.star.cs.uiowa.edu
space	AG01_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.74814295769 seconds
cpu usage	3.886945766
max memory	1.97206016E8

stage attributes

key	value
output-size	6497
starexec-result	YES

output

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


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YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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) QReductionProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) UsableRulesProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) QReductionProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (18) YES


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

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

The set Q consists of the following terms:

   f(x0, c(x0), c(x1))
   f(s(x0), x1, x2)
   f(c(x0), x0, x1)
   g(x0, x1)


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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:

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   f(x0, c(x0), c(x1))
   f(s(x0), x1, x2)
   f(c(x0), x0, x1)
   g(x0, x1)

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 2 SCCs with 1 less node.
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(4)
Complex Obligation (AND)

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(5)
Obligation:
Q DP problem:

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