YES
proof of SK90_2.25.trs
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 61 ms]
(2) QTRS
(3) RisEmptyProof [EQUIVALENT, 0 ms]
(4) YES


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

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

   fib(0) -> 0
   fib(s(0)) -> s(0)
   fib(s(s(x))) -> +(fib(s(x)), fib(x))
   +(x, 0) -> x
   +(x, s(y)) -> s(+(x, y))

Q is empty.

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

(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Quasi precedence:
fib_1 > +_2 > s_1
0 > s_1


Status:
fib_1: multiset status
0: multiset status
s_1: multiset status
+_2: [2,1]

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   fib(0) -> 0
   fib(s(0)) -> s(0)
   fib(s(s(x))) -> +(fib(s(x)), fib(x))
   +(x, 0) -> x
   +(x, s(y)) -> s(+(x, y))




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

(2)
Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.

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

(3) RisEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
----------------------------------------

(4)
YES