TRS Innermost pair #487093051

details

property	value
status	complete
benchmark	Ex24_Luc06_iGM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n144.star.cs.uiowa.edu
space	Transformed_CSR_innermost_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.9395 seconds
cpu usage	4.38463
user time	4.20748
system time	0.177148
max virtual memory	1.8412788E7
max residence set size	270108.0

stage attributes

key	value
starexec-result	YES

output

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 28 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) MRRProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) TransformationProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) UsableRulesProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) TransformationProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) UsableRulesProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) TransformationProof [EQUIVALENT, 0 ms]
        (24) QDP
        (25) UsableRulesProof [EQUIVALENT, 0 ms]
        (26) QDP
        (27) TransformationProof [EQUIVALENT, 0 ms]
        (28) QDP
        (29) UsableRulesProof [EQUIVALENT, 0 ms]
        (30) QDP
        (31) TransformationProof [EQUIVALENT, 0 ms]
        (32) QDP
        (33) DependencyGraphProof [EQUIVALENT, 0 ms]
        (34) TRUE


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

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

   active(f(b, X, c)) -> mark(f(X, c, X))
   active(c) -> mark(b)
   mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))
   mark(b) -> active(b)
   mark(c) -> active(c)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

The set Q consists of the following terms:

   active(f(b, x0, c))
   active(c)
   mark(f(x0, x1, x2))
   mark(b)
   mark(c)
   f(mark(x0), x1, x2)
   f(x0, mark(x1), x2)
   f(x0, x1, mark(x2))
   f(active(x0), x1, x2)
   f(x0, active(x1), x2)
   f(x0, x1, active(x2))


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

(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:

   ACTIVE(f(b, X, c)) -> MARK(f(X, c, X))
   ACTIVE(f(b, X, c)) -> F(X, c, X)
   ACTIVE(c) -> MARK(b)
   MARK(f(X1, X2, X3)) -> ACTIVE(f(X1, mark(X2), X3))
   MARK(f(X1, X2, X3)) -> F(X1, mark(X2), X3)
   MARK(f(X1, X2, X3)) -> MARK(X2)
   MARK(b) -> ACTIVE(b)
   MARK(c) -> ACTIVE(c)
   F(mark(X1), X2, X3) -> F(X1, X2, X3)
   F(X1, mark(X2), X3) -> F(X1, X2, X3)
   F(X1, X2, mark(X3)) -> F(X1, X2, X3)

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