TRS Standard pair #487072298

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	n192.star.cs.uiowa.edu
space	Secret_07_TRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	32.4018 seconds
cpu usage	107.046
user time	103.039
system time	4.00689
max virtual memory	1.9393528E7
max residence set size	6180976.0

stage attributes

key	value
starexec-result	YES

output

YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 24 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPOrderProof [EQUIVALENT, 9 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, 25 ms]
        (18) QDP
        (19) QDPBoundsTAProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) PisEmptyProof [EQUIVALENT, 0 ms]
        (22) YES
    (23) QDP
        (24) QDPOrderProof [EQUIVALENT, 15 ms]
        (25) QDP
        (26) TransformationProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) QDPOrderProof [EQUIVALENT, 2017 ms]
        (29) QDP
        (30) PisEmptyProof [EQUIVALENT, 0 ms]
        (31) YES


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

(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.

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

(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:

   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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(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 11 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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