TRS Inner 89993 pair #381733927

details

property	value
status	complete
benchmark	Ex4_4_Luc96b_iGM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n045.star.cs.uiowa.edu
space	Transformed_CSR_innermost_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.85393095016 seconds
cpu usage	4.231766722
max memory	2.64548352E8

stage attributes

key	value
output-size	14765
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 3 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) QDPOrderProof [EQUIVALENT, 12 ms]
        (21) QDP
        (22) DependencyGraphProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) UsableRulesProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) QReductionProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (29) YES


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

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

   active(f(g(X), Y)) -> mark(f(X, f(g(X), Y)))
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(g(X)) -> active(g(mark(X)))
   f(mark(X1), X2) -> f(X1, X2)
   f(X1, mark(X2)) -> f(X1, X2)
   f(active(X1), X2) -> f(X1, X2)
   f(X1, active(X2)) -> f(X1, X2)
   g(mark(X)) -> g(X)
   g(active(X)) -> g(X)

The set Q consists of the following terms:

   active(f(g(x0), x1))
   mark(f(x0, x1))
   mark(g(x0))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))
   g(mark(x0))
   g(active(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:

   ACTIVE(f(g(X), Y)) -> MARK(f(X, f(g(X), Y)))
   ACTIVE(f(g(X), Y)) -> F(X, f(g(X), Y))
   MARK(f(X1, X2)) -> ACTIVE(f(mark(X1), X2))
   MARK(f(X1, X2)) -> F(mark(X1), X2)
   MARK(f(X1, X2)) -> MARK(X1)
   MARK(g(X)) -> ACTIVE(g(mark(X)))
   MARK(g(X)) -> G(mark(X))
   MARK(g(X)) -> MARK(X)
   F(mark(X1), X2) -> F(X1, X2)
   F(X1, mark(X2)) -> F(X1, X2)
   F(active(X1), X2) -> F(X1, X2)
   F(X1, active(X2)) -> F(X1, X2)
   G(mark(X)) -> G(X)
   G(active(X)) -> G(X)

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