TRS Standard pair #615300005

details

property	value
status	complete
benchmark	Ex4_7_56_Bor03_GM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n169.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE23
configuration	standard
runtime (wallclock)	4.54363512993 seconds
cpu usage	15.138962329
max memory	8.974336E8

stage attributes

key	value
output-size	25710
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# AProVE Commit ID: 363435ff5b19173ec8a83b45cfdec073678dabcc jan-christoph 20230815 unpublished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) QDPOrderProof [EQUIVALENT, 81 ms]
(4) QDP
(5) QDPOrderProof [EQUIVALENT, 87 ms]
(6) QDP
(7) QDPOrderProof [EQUIVALENT, 74 ms]
(8) QDP
(9) QDPOrderProof [EQUIVALENT, 28 ms]
(10) QDP
(11) QDPOrderProof [EQUIVALENT, 11 ms]
(12) QDP
(13) DependencyGraphProof [EQUIVALENT, 3 ms]
(14) AND
    (15) QDP
        (16) QDPOrderProof [EQUIVALENT, 20 ms]
        (17) QDP
        (18) PisEmptyProof [EQUIVALENT, 0 ms]
        (19) YES
    (20) QDP
        (21) QDPOrderProof [EQUIVALENT, 8 ms]
        (22) QDP
        (23) QDPOrderProof [EQUIVALENT, 13 ms]
        (24) QDP
        (25) DependencyGraphProof [EQUIVALENT, 0 ms]
        (26) QDP
        (27) UsableRulesProof [EQUIVALENT, 0 ms]
        (28) QDP
        (29) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (30) YES


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

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

   a__from(X) -> cons(mark(X), from(s(X)))
   a__after(0, XS) -> mark(XS)
   a__after(s(N), cons(X, XS)) -> a__after(mark(N), mark(XS))
   mark(from(X)) -> a__from(mark(X))
   mark(after(X1, X2)) -> a__after(mark(X1), mark(X2))
   mark(cons(X1, X2)) -> cons(mark(X1), X2)
   mark(s(X)) -> s(mark(X))
   mark(0) -> 0
   a__from(X) -> from(X)
   a__after(X1, X2) -> after(X1, X2)

Q is empty.

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

(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:

   A__FROM(X) -> MARK(X)
   A__AFTER(0, XS) -> MARK(XS)
   A__AFTER(s(N), cons(X, XS)) -> A__AFTER(mark(N), mark(XS))
   A__AFTER(s(N), cons(X, XS)) -> MARK(N)
   A__AFTER(s(N), cons(X, XS)) -> MARK(XS)
   MARK(from(X)) -> A__FROM(mark(X))
   MARK(from(X)) -> MARK(X)
   MARK(after(X1, X2)) -> A__AFTER(mark(X1), mark(X2))
   MARK(after(X1, X2)) -> MARK(X1)
   MARK(after(X1, X2)) -> MARK(X2)
   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(s(X)) -> MARK(X)

The TRS R consists of the following rules:

   a__from(X) -> cons(mark(X), from(s(X)))
   a__after(0, XS) -> mark(XS)
   a__after(s(N), cons(X, XS)) -> a__after(mark(N), mark(XS))
   mark(from(X)) -> a__from(mark(X))
   mark(after(X1, X2)) -> a__after(mark(X1), mark(X2))
   mark(cons(X1, X2)) -> cons(mark(X1), X2)
   mark(s(X)) -> s(mark(X))
   mark(0) -> 0

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