TRS Standard pair #615297940

details

property	value
status	complete
benchmark	2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n026.star.cs.uiowa.edu
space	Secret_07_TRS

run statistics

property	value
solver	AProVE23
configuration	standard
runtime (wallclock)	36.5808758736 seconds
cpu usage	114.934894939
max memory	5.99052288E9

stage attributes

key	value
output-size	13805
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# AProVE Commit ID: 363435ff5b19173ec8a83b45cfdec073678dabcc jan-christoph 20230815 unpublished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) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPOrderProof [EQUIVALENT, 22 ms]
        (9) QDP
        (10) PisEmptyProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) UsableRulesProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) TransformationProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QDPBoundsTAProof [EQUIVALENT, 22 ms]
        (18) QDP
        (19) QDPBoundsTAProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) PisEmptyProof [EQUIVALENT, 0 ms]
        (22) YES
    (23) QDP
        (24) QDPOrderProof [EQUIVALENT, 23 ms]
        (25) QDP
        (26) TransformationProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) QDPOrderProof [EQUIVALENT, 2201 ms]
        (29) QDP
        (30) PisEmptyProof [EQUIVALENT, 0 ms]
        (31) YES


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

   h(x, x) -> h(a, b)
   g(g(x, a), y) -> g(g(a, y), g(a, x))
   f(g(x, y)) -> g(g(f(f(y)), h(a, a)), x)
   h(h(f(f(x)), y), h(z, v)) -> h(h(f(z), f(f(f(y)))), h(v, x))

Q is empty.

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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:

   H(x, x) -> H(a, b)
   G(g(x, a), y) -> G(g(a, y), g(a, x))
   G(g(x, a), y) -> G(a, y)
   G(g(x, a), y) -> G(a, x)
   F(g(x, y)) -> G(g(f(f(y)), h(a, a)), x)
   F(g(x, y)) -> G(f(f(y)), h(a, a))
   F(g(x, y)) -> F(f(y))
   F(g(x, y)) -> F(y)
   F(g(x, y)) -> H(a, a)
   H(h(f(f(x)), y), h(z, v)) -> H(h(f(z), f(f(f(y)))), h(v, x))
   H(h(f(f(x)), y), h(z, v)) -> H(f(z), f(f(f(y))))
   H(h(f(f(x)), y), h(z, v)) -> F(z)
   H(h(f(f(x)), y), h(z, v)) -> F(f(f(y)))
   H(h(f(f(x)), y), h(z, v)) -> F(f(y))
   H(h(f(f(x)), y), h(z, v)) -> F(y)
   H(h(f(f(x)), y), h(z, v)) -> H(v, x)

The TRS R consists of the following rules:

   h(x, x) -> h(a, b)
   g(g(x, a), y) -> g(g(a, y), g(a, x))
   f(g(x, y)) -> g(g(f(f(y)), h(a, a)), x)
   h(h(f(f(x)), y), h(z, v)) -> h(h(f(z), f(f(f(y)))), h(v, x))

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