TRS Standard pair #516963485

details

property	value
status	complete
benchmark	Ex4_7_37_Bor03_Z.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n172.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.48339796066 seconds
cpu usage	6.318079418
max memory	4.85707776E8

stage attributes

key	value
output-size	11253
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished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) QDPOrderProof [EQUIVALENT, 8 ms]
        (12) QDP
        (13) PisEmptyProof [EQUIVALENT, 0 ms]
        (14) YES
    (15) QDP
        (16) UsableRulesProof [EQUIVALENT, 0 ms]
        (17) QDP
        (18) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (19) YES
    (20) QDP
        (21) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (22) YES


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

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

   from(X) -> cons(X, n__from(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)
   zWquot(X1, X2) -> n__zWquot(X1, X2)
   activate(n__from(X)) -> from(X)
   activate(n__zWquot(X1, X2)) -> zWquot(X1, 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)) -> 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(X)
   ACTIVATE(n__zWquot(X1, X2)) -> ZWQUOT(X1, X2)

The TRS R consists of the following rules:

   from(X) -> cons(X, n__from(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)
   zWquot(X1, X2) -> n__zWquot(X1, X2)
   activate(n__from(X)) -> from(X)

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job log

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