TRS Standard pair #615297275

details

property	value
status	complete
benchmark	z19.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n175.star.cs.uiowa.edu
space	Zantema_05

run statistics

property	value
solver	AProVE23
configuration	standard
runtime (wallclock)	14.4630479813 seconds
cpu usage	33.21793653
max memory	1.550573568E9

stage attributes

key	value
output-size	19736
starexec-result	"YES"

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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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# AProVE Commit ID: 363435ff5b19173ec8a83b45cfdec073678dabcc jan-christoph 20230815 unpublished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) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) MRRProof [EQUIVALENT, 0 ms]
(4) QDP
(5) SemLabProof [SOUND, 197 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) QDP
(9) QDPOrderProof [EQUIVALENT, 15 ms]
(10) QDP
(11) QDPOrderProof [EQUIVALENT, 12 ms]
(12) QDP
(13) DependencyGraphProof [EQUIVALENT, 0 ms]
(14) AND
    (15) QDP
        (16) QDPOrderProof [EQUIVALENT, 7 ms]
        (17) QDP
        (18) QDPOrderProof [EQUIVALENT, 0 ms]
        (19) QDP
        (20) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
        (21) QDP
        (22) PisEmptyProof [EQUIVALENT, 0 ms]
        (23) YES
    (24) QDP
        (25) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
        (26) QDP
        (27) PisEmptyProof [EQUIVALENT, 0 ms]
        (28) YES


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

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

   f(f(a, a), x) -> f(a, f(b, f(a, x)))
   f(x, f(y, z)) -> f(f(x, 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.
----------------------------------------

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(f(a, a), x) -> F(a, f(b, f(a, x)))
   F(f(a, a), x) -> F(b, f(a, x))
   F(f(a, a), x) -> F(a, x)
   F(x, f(y, z)) -> F(f(x, y), z)
   F(x, f(y, z)) -> F(x, y)

The TRS R consists of the following rules:

   f(f(a, a), x) -> f(a, f(b, f(a, x)))
   f(x, f(y, z)) -> f(f(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(3) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   F(f(a, a), x) -> F(b, f(a, x))
   F(f(a, a), x) -> F(a, x)


Used ordering: Polynomial interpretation [POLO]:

   POL(F(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(a) = 2
   POL(b) = 0
   POL(f(x_1, x_2)) = x_1 + x_2


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