TRS Innermost pair #487524019

details

property	value
status	complete
benchmark	#4.33.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n148.star.cs.uiowa.edu
space	AG01_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.83018279076 seconds
cpu usage	4.181820963
max memory	2.52104704E8

stage attributes

key	value
output-size	9930
starexec-result	YES

output

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


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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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 27 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) UsableRulesReductionPairsProof [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) QDPOrderProof [EQUIVALENT, 26 ms]
        (20) QDP
        (21) PisEmptyProof [EQUIVALENT, 0 ms]
        (22) YES


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

   sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
   sum(cons(0, x), y) -> sum(x, y)
   sum(nil, y) -> y
   weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
   weight(cons(n, nil)) -> n

The set Q consists of the following terms:

   sum(cons(s(x0), x1), cons(x2, x3))
   sum(cons(0, x0), x1)
   sum(nil, x0)
   weight(cons(x0, cons(x1, x2)))
   weight(cons(x0, nil))


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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SUM(cons(s(n), x), cons(m, y)) -> SUM(cons(n, x), cons(s(m), y))
   SUM(cons(0, x), y) -> SUM(x, y)
   WEIGHT(cons(n, cons(m, x))) -> WEIGHT(sum(cons(n, cons(m, x)), cons(0, x)))
   WEIGHT(cons(n, cons(m, x))) -> SUM(cons(n, cons(m, x)), cons(0, x))

The TRS R consists of the following rules:

   sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
   sum(cons(0, x), y) -> sum(x, y)
   sum(nil, y) -> y
   weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
   weight(cons(n, nil)) -> n

The set Q consists of the following terms:

   sum(cons(s(x0), x1), cons(x2, x3))
   sum(cons(0, x0), x1)
   sum(nil, x0)
   weight(cons(x0, cons(x1, x2)))
   weight(cons(x0, nil))

We have to consider all minimal (P,Q,R)-chains.
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(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
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(4)

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