TRS Standard pair #487069703

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	n073.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.21771 seconds
cpu usage	5.37714
user time	5.15278
system time	0.22436
max virtual memory	1.8544088E7
max residence set size	366532.0

stage attributes

key	value
starexec-result	YES

output

YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 7 ms]
        (17) QDP
        (18) QDPOrderProof [EQUIVALENT, 0 ms]
        (19) QDP
        (20) UsableRulesProof [EQUIVALENT, 0 ms]
        (21) QDP
        (22) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (23) YES
    (24) QDP
        (25) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (26) 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))
   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)))

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