TRS Inner 89993 pair #381733518

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	n005.star.cs.uiowa.edu
space	AG01_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.62711310387 seconds
cpu usage	3.668511929
max memory	2.40078848E8

stage attributes

key	value
output-size	6599
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QReductionProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) MRRProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) PisEmptyProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QReductionProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) MRRProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) PisEmptyProof [EQUIVALENT, 0 ms]
        (22) 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))

The set Q consists of the following terms:

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


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

(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(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.
----------------------------------------

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

(4)
Complex Obligation (AND)

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

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

   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)))

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