TRS Standard pair #615300495

details

property	value
status	complete
benchmark	Ex16_Luc06_iGM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n157.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE23
configuration	standard
runtime (wallclock)	2.25652694702 seconds
cpu usage	6.404738185
max memory	3.28544256E8

stage attributes

key	value
output-size	11637
starexec-result	"YES"

output

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


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YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 363435ff5b19173ec8a83b45cfdec073678dabcc jan-christoph 20230815 unpublished


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) MRRProof [EQUIVALENT, 20 ms]
        (12) QDP
        (13) TransformationProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) TransformationProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) DependencyGraphProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) UsableRulesProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) TransformationProof [EQUIVALENT, 0 ms]
        (24) QDP
        (25) DependencyGraphProof [EQUIVALENT, 0 ms]
        (26) TRUE


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

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

Q is empty.

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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:

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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