TRS Innermost pair #487092983

details

property	value
status	complete
benchmark	#4.37a.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n149.star.cs.uiowa.edu
space	AG01_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.72161 seconds
cpu usage	3.93564
user time	3.74789
system time	0.187749
max virtual memory	1.854204E7
max residence set size	258312.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRSRRRProof [EQUIVALENT, 57 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 1 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) MRRProof [EQUIVALENT, 0 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))
   g(s(f(x))) -> g(f(x))

The set Q consists of the following terms:

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


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

(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(c(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(f(x_1)) = 2*x_1
   POL(g(x_1)) = 2*x_1
   POL(s(x_1)) = 1 + x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   g(s(f(x))) -> g(f(x))




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

(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)))
   g(s(f(x0)))


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

(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))

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