TRS Standard pair #487067638

details

property	value
status	complete
benchmark	#3.47.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n191.star.cs.uiowa.edu
space	AG01

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.64785 seconds
cpu usage	3.66371
user time	3.496
system time	0.167706
max virtual memory	1.8408912E7
max residence set size	241320.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (12) YES


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

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

   f(x, c(y)) -> f(x, s(f(y, y)))
   f(s(x), y) -> f(x, s(c(y)))

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:

   F(x, c(y)) -> F(x, s(f(y, y)))
   F(x, c(y)) -> F(y, y)
   F(s(x), y) -> F(x, s(c(y)))

The TRS R consists of the following rules:

   f(x, c(y)) -> f(x, s(f(y, y)))
   f(s(x), y) -> f(x, s(c(y)))

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:

   F(s(x), y) -> F(x, s(c(y)))

The TRS R consists of the following rules:

   f(x, c(y)) -> f(x, s(f(y, y)))
   f(s(x), y) -> f(x, s(c(y)))

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.
----------------------------------------

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(s(x), y) -> F(x, s(c(y)))

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

(8) QDPSizeChangeProof (EQUIVALENT)

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