TRS Inner 89993 pair #381733893

details

property	value
status	complete
benchmark	Ex9_Luc06_iGM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n026.star.cs.uiowa.edu
space	Transformed_CSR_innermost_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.91344594955 seconds
cpu usage	4.372645053
max memory	2.58260992E8

stage attributes

key	value
output-size	21399
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) MRRProof [EQUIVALENT, 27 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(a, X, X)) -> mark(f(X, b, b))
   active(b) -> mark(a)
   mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))
   mark(a) -> active(a)
   mark(b) -> active(b)
   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(a, x0, x0))
   active(b)
   mark(f(x0, x1, x2))
   mark(a)
   mark(b)
   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))


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

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