SRS Standard pair #487512647

details

property	value
status	complete
benchmark	size-12-alpha-3-num-275.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n116.star.cs.uiowa.edu
space	Waldmann_07_size12

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	5.22114992142 seconds
cpu usage	16.889819494
max memory	1.988562944E9

stage attributes

key	value
output-size	7233
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 17 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 6 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) MRRProof [EQUIVALENT, 8 ms]
        (11) QDP
        (12) MRRProof [EQUIVALENT, 7 ms]
        (13) QDP
        (14) PisEmptyProof [EQUIVALENT, 0 ms]
        (15) YES
    (16) QDP
        (17) UsableRulesProof [EQUIVALENT, 2 ms]
        (18) QDP
        (19) QDPOrderProof [EQUIVALENT, 59 ms]
        (20) QDP
        (21) PisEmptyProof [EQUIVALENT, 0 ms]
        (22) YES


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

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

   a(x1) -> x1
   a(b(x1)) -> c(b(x1))
   a(c(c(x1))) -> c(c(a(a(x1))))

Q is empty.

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

(1) QTRS Reverse (EQUIVALENT)
We applied the QTRS Reverse Processor [REVERSE].
----------------------------------------

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

   a(x1) -> x1
   b(a(x1)) -> b(c(x1))
   c(c(a(x1))) -> a(a(c(c(x1))))

Q is empty.

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

(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:

   B(a(x1)) -> B(c(x1))
   B(a(x1)) -> C(x1)
   C(c(a(x1))) -> A(a(c(c(x1))))
   C(c(a(x1))) -> A(c(c(x1)))
   C(c(a(x1))) -> C(c(x1))
   C(c(a(x1))) -> C(x1)

The TRS R consists of the following rules:

   a(x1) -> x1
   b(a(x1)) -> b(c(x1))
   c(c(a(x1))) -> a(a(c(c(x1))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.
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