TRS Standard pair #615296771

details

property	value
status	complete
benchmark	gen-22.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n178.star.cs.uiowa.edu
space	Secret_06_TRS

run statistics

property	value
solver	AProVE23
configuration	standard
runtime (wallclock)	2.40374588966 seconds
cpu usage	6.710030263
max memory	3.41389312E8

stage attributes

key	value
output-size	6944
starexec-result	"YES"

output

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


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


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# AProVE Commit ID: 363435ff5b19173ec8a83b45cfdec073678dabcc jan-christoph 20230815 unpublished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, 20 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) UsableRulesReductionPairsProof [EQUIVALENT, 17 ms]
        (9) QDP
        (10) PisEmptyProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) QDPOrderProof [EQUIVALENT, 3 ms]
        (14) QDP
        (15) PisEmptyProof [EQUIVALENT, 0 ms]
        (16) YES


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

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

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:

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

The TRS R consists of the following rules:

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

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 2 SCCs with 9 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   B(a, b(c(z, x, y), a)) -> C(y, z, a)
   C(f(c(a, y, a)), x, z) -> B(y, b(x, a))

The TRS R consists of the following rules:

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

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