TRS Innermost pair #487524631

details

property	value
status	complete
benchmark	cade13.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n156.star.cs.uiowa.edu
space	Mixed_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.61600899696 seconds
cpu usage	6.668765857
max memory	3.84155648E8

stage attributes

key	value
output-size	18183
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 11 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) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) UsableRulesProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) QReductionProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (18) YES
    (19) QDP
        (20) UsableRulesProof [EQUIVALENT, 0 ms]
        (21) QDP
        (22) QReductionProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) QDPQMonotonicMRRProof [EQUIVALENT, 26 ms]
        (25) QDP
        (26) NonInfProof [EQUIVALENT, 51 ms]
        (27) QDP
        (28) DependencyGraphProof [EQUIVALENT, 0 ms]
        (29) TRUE


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

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

   div(x, s(y)) -> d(x, s(y), 0)
   d(x, s(y), z) -> cond(ge(x, z), x, y, z)
   cond(true, x, y, z) -> s(d(x, s(y), plus(s(y), z)))
   cond(false, x, y, z) -> 0
   ge(u, 0) -> true
   ge(0, s(v)) -> false
   ge(s(u), s(v)) -> ge(u, v)
   plus(n, 0) -> n
   plus(n, s(m)) -> s(plus(n, m))
   plus(plus(n, m), u) -> plus(n, plus(m, u))

The set Q consists of the following terms:

   div(x0, s(x1))
   d(x0, s(x1), x2)
   cond(true, x0, x1, x2)
   cond(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   plus(x0, 0)
   plus(x0, s(x1))
   plus(plus(x0, x1), x2)


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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:

   DIV(x, s(y)) -> D(x, s(y), 0)
   D(x, s(y), z) -> COND(ge(x, z), x, y, z)
   D(x, s(y), z) -> GE(x, z)
   COND(true, x, y, z) -> D(x, s(y), plus(s(y), z))
   COND(true, x, y, z) -> PLUS(s(y), z)
   GE(s(u), s(v)) -> GE(u, v)
   PLUS(n, s(m)) -> PLUS(n, m)
   PLUS(plus(n, m), u) -> PLUS(n, plus(m, u))
   PLUS(plus(n, m), u) -> PLUS(m, u)

The TRS R consists of the following rules:

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