TRS Innermost pair #487524625

details

property	value
status	complete
benchmark	innermost5.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n107.star.cs.uiowa.edu
space	Mixed_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.31262898445 seconds
cpu usage	4.481762849
max memory	3.15174912E8

stage attributes

key	value
output-size	11668
starexec-result	YES

output

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


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


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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 12 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, 14 ms]
        (25) QDP
        (26) UsableRulesProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) QDPOrderProof [EQUIVALENT, 0 ms]
        (29) QDP
        (30) PisEmptyProof [EQUIVALENT, 0 ms]
        (31) YES


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

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

   f(s(x), y, b) -> f(g(h(x)), y, i(y))
   g(h(x)) -> g(x)
   g(s(x)) -> s(x)
   g(0) -> s(0)
   h(0) -> a
   i(0) -> b
   i(s(y)) -> i(y)

The set Q consists of the following terms:

   f(s(x0), x1, b)
   g(h(x0))
   g(s(x0))
   g(0)
   h(0)
   i(0)
   i(s(x0))


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

(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(s(x), y, b) -> F(g(h(x)), y, i(y))
   F(s(x), y, b) -> G(h(x))
   F(s(x), y, b) -> H(x)
   F(s(x), y, b) -> I(y)
   G(h(x)) -> G(x)
   I(s(y)) -> I(y)

The TRS R consists of the following rules:

   f(s(x), y, b) -> f(g(h(x)), y, i(y))
   g(h(x)) -> g(x)
   g(s(x)) -> s(x)
   g(0) -> s(0)
   h(0) -> a
   i(0) -> b
   i(s(y)) -> i(y)

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