TRS Standard pair #516965770

details

property	value
status	complete
benchmark	pair2hard.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n028.star.cs.uiowa.edu
space	Endrullis_06

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	39.9134128094 seconds
cpu usage	104.921855957
max memory	2.894548992E9

stage attributes

key	value
output-size	7910
starexec-result	YES

output

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


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


YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished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) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) PisEmptyProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) QDPOrderProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) TransformationProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QDPOrderProof [EQUIVALENT, 19.9 s]
        (18) QDP
        (19) PisEmptyProof [EQUIVALENT, 0 ms]
        (20) YES


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

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

   p(a(x0), p(b(a(x1)), x2)) -> p(x1, p(a(b(a(x1))), x2))
   a(b(a(x0))) -> b(a(b(x0)))

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:

   P(a(x0), p(b(a(x1)), x2)) -> P(x1, p(a(b(a(x1))), x2))
   P(a(x0), p(b(a(x1)), x2)) -> P(a(b(a(x1))), x2)
   P(a(x0), p(b(a(x1)), x2)) -> A(b(a(x1)))
   A(b(a(x0))) -> A(b(x0))

The TRS R consists of the following rules:

   p(a(x0), p(b(a(x1)), x2)) -> p(x1, p(a(b(a(x1))), x2))
   a(b(a(x0))) -> b(a(b(x0)))

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

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
----------------------------------------

(4)
Complex Obligation (AND)

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

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

   A(b(a(x0))) -> A(b(x0))

The TRS R consists of the following rules:

   p(a(x0), p(b(a(x1)), x2)) -> p(x1, p(a(b(a(x1))), x2))
   a(b(a(x0))) -> b(a(b(x0)))

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

(6) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

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