(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, fib1(Y, add(X, Y)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, XS)
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
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:
FIB(N) → SEL(N, fib1(s(0), s(0)))
FIB(N) → FIB1(s(0), s(0))
FIB1(X, Y) → FIB1(Y, add(X, Y))
FIB1(X, Y) → ADD(X, Y)
ADD(s(X), Y) → ADD(X, Y)
SEL(s(N), cons(X, XS)) → SEL(N, XS)
The TRS R consists of the following rules:
fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, fib1(Y, add(X, Y)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, XS)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 3 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
SEL(s(N), cons(X, XS)) → SEL(N, XS)
The TRS R consists of the following rules:
fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, fib1(Y, add(X, Y)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, XS)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- SEL(s(N), cons(X, XS)) → SEL(N, XS)
The graph contains the following edges 1 > 1, 2 > 2
(7) YES
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ADD(s(X), Y) → ADD(X, Y)
The TRS R consists of the following rules:
fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, fib1(Y, add(X, Y)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, XS)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- ADD(s(X), Y) → ADD(X, Y)
The graph contains the following edges 1 > 1, 2 >= 2
(10) YES
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FIB1(X, Y) → FIB1(Y, add(X, Y))
The TRS R consists of the following rules:
fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, fib1(Y, add(X, Y)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, XS)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) NonLoopProof (EQUIVALENT transformation)
By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 0,
σ' = [ ], and μ' = [x1 / x0, x0 / add(x1, x0)] on the rule
FIB1(x0, add(x1, x0))[ ]n[ ] → FIB1(x0, add(x1, x0))[ ]n[x1 / x0, x0 / add(x1, x0)]
This rule is correct for the QDP as the following derivation shows:
intermediate steps: Equivalent (Simplify mu) - Instantiate mu - Instantiation
FIB1(X, Y)[ ]n[ ] → FIB1(Y, add(X, Y))[ ]n[ ]
by OriginalRule from TRS P
(13) NO