TRS Standard pair #487066988

details

property	value
status	complete
benchmark	#4.37.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n175.star.cs.uiowa.edu
space	Strategy_removed_AG01

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.71253 seconds
cpu usage	3.67066
user time	3.50534
system time	0.165321
max virtual memory	1.8675396E7
max residence set size	249848.0

stage attributes

key	value
starexec-result	YES

output

YES
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) AAECC Innermost [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) MRRProof [EQUIVALENT, 9 ms]
        (13) QDP
        (14) PisEmptyProof [EQUIVALENT, 0 ms]
        (15) YES
    (16) QDP
        (17) UsableRulesProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QReductionProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) MRRProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) PisEmptyProof [EQUIVALENT, 0 ms]
        (24) YES


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

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

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

Q is empty.

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

(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none

The TRS R 2 is 
   f(c(s(x), y)) -> f(c(x, s(y)))
   g(c(x, s(y))) -> g(c(s(x), y))

The signature Sigma is {f_1, g_1}
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   f(c(s(x0), x1))
   g(c(x0, s(x1)))


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

(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   f(c(s(x0), x1))
   g(c(x0, s(x1)))

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

(5) DependencyGraphProof (EQUIVALENT)

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