TRS Innermost pair #487092979

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	n150.star.cs.uiowa.edu
space	AG01_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.83771 seconds
cpu usage	4.17036
user time	3.9977
system time	0.172669
max virtual memory	1.841096E7
max residence set size	251332.0

stage attributes

key	value
starexec-result	YES

output

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 29 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)
Complex Obligation (AND)

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(5)
Obligation:

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