TRS Stand 20472 pair #381716252

details

property	value
status	complete
benchmark	Ex4_7_37_Bor03_FR.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n079.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.12221002579 seconds
cpu usage	5.061609891
max memory	2.95010304E8

stage attributes

key	value
output-size	14786
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) QDPOrderProof [EQUIVALENT, 0 ms]
        (12) QDP
        (13) PisEmptyProof [EQUIVALENT, 0 ms]
        (14) YES
    (15) QDP
        (16) QDPOrderProof [EQUIVALENT, 11 ms]
        (17) QDP
        (18) UsableRulesProof [EQUIVALENT, 0 ms]
        (19) QDP
        (20) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (21) YES
    (22) QDP
        (23) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (24) YES


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

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

   from(X) -> cons(X, n__from(n__s(X)))
   sel(0, cons(X, XS)) -> X
   sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
   minus(X, 0) -> 0
   minus(s(X), s(Y)) -> minus(X, Y)
   quot(0, s(Y)) -> 0
   quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
   zWquot(XS, nil) -> nil
   zWquot(nil, XS) -> nil
   zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), n__zWquot(activate(XS), activate(YS)))
   from(X) -> n__from(X)
   s(X) -> n__s(X)
   zWquot(X1, X2) -> n__zWquot(X1, X2)
   activate(n__from(X)) -> from(activate(X))
   activate(n__s(X)) -> s(activate(X))
   activate(n__zWquot(X1, X2)) -> zWquot(activate(X1), activate(X2))
   activate(X) -> X

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:

   SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS))
   SEL(s(N), cons(X, XS)) -> ACTIVATE(XS)
   MINUS(s(X), s(Y)) -> MINUS(X, Y)
   QUOT(s(X), s(Y)) -> S(quot(minus(X, Y), s(Y)))
   QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))
   QUOT(s(X), s(Y)) -> MINUS(X, Y)
   ZWQUOT(cons(X, XS), cons(Y, YS)) -> QUOT(X, Y)
   ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(XS)
   ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS)
   ACTIVATE(n__from(X)) -> FROM(activate(X))
   ACTIVATE(n__from(X)) -> ACTIVATE(X)
   ACTIVATE(n__s(X)) -> S(activate(X))
   ACTIVATE(n__s(X)) -> ACTIVATE(X)
   ACTIVATE(n__zWquot(X1, X2)) -> ZWQUOT(activate(X1), activate(X2))
   ACTIVATE(n__zWquot(X1, X2)) -> ACTIVATE(X1)
   ACTIVATE(n__zWquot(X1, X2)) -> ACTIVATE(X2)

The TRS R consists of the following rules:

   from(X) -> cons(X, n__from(n__s(X)))
   sel(0, cons(X, XS)) -> X
   sel(s(N), cons(X, XS)) -> sel(N, activate(XS))

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