TRS Standard pair #487066578

details

property	value
status	complete
benchmark	ex1.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n185.star.cs.uiowa.edu
space	AProVE_10

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	11.867 seconds
cpu usage	24.7511
user time	24.2069
system time	0.544114
max virtual memory	3.7541412E7
max residence set size	681212.0

stage attributes

key	value
starexec-result	NO

output

NO
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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 disproven:

(0) QTRS
(1) AAECC Innermost [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) MNOCProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) NonLoopProof [COMPLETE, 81 ms]
        (18) NO


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

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

   f(x, y) -> cond(lt(x, y), x, y)
   cond(tt, x, y) -> f(s(x), s(y))
   lt(0, y) -> tt
   lt(s(x), s(y)) -> lt(x, y)

Q is empty.

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

(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   lt(0, y) -> tt
   lt(s(x), s(y)) -> lt(x, y)

The TRS R 2 is 
   f(x, y) -> cond(lt(x, y), x, y)
   cond(tt, x, y) -> f(s(x), s(y))

The signature Sigma is {f_2, cond_3}
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(x, y) -> cond(lt(x, y), x, y)
   cond(tt, x, y) -> f(s(x), s(y))
   lt(0, y) -> tt
   lt(s(x), s(y)) -> lt(x, y)

The set Q consists of the following terms:

   f(x0, x1)
   cond(tt, x0, x1)
   lt(0, x0)
   lt(s(x0), s(x1))


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

(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(x, y) -> COND(lt(x, y), x, y)
   F(x, y) -> LT(x, y)
   COND(tt, x, y) -> F(s(x), s(y))
   LT(s(x), s(y)) -> LT(x, y)

The TRS R consists of the following rules:

   f(x, y) -> cond(lt(x, y), x, y)
   cond(tt, x, y) -> f(s(x), s(y))
   lt(0, y) -> tt
   lt(s(x), s(y)) -> lt(x, y)

The set Q consists of the following terms:

   f(x0, x1)

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