(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

f(X) → if(X, c, n__f(true))
if(true, X, Y) → X
if(false, X, Y) → activate(Y)
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → X

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:

F(X) → IF(X, c, n__f(true))
IF(false, X, Y) → ACTIVATE(Y)
ACTIVATE(n__f(X)) → F(X)

The TRS R consists of the following rules:

f(X) → if(X, c, n__f(true))
if(true, X, Y) → X
if(false, X, Y) → activate(Y)
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(X) → IF(X, c, n__f(true))
IF(false, X, Y) → ACTIVATE(Y)
ACTIVATE(n__f(X)) → F(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
F(x0, x1)  =  F(x0)
IF(x0, x1, x2, x3)  =  IF(x0, x1, x3)
ACTIVATE(x0, x1)  =  ACTIVATE(x0)

Tags:
F has argument tags [3,2] and root tag 3
IF has argument tags [0,3,4,3] and root tag 1
ACTIVATE has argument tags [7,1] and root tag 2

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(ACTIVATE(x1)) = x1   
POL(F(x1)) = x1   
POL(IF(x1, x2, x3)) = x1 + x3   
POL(c) = 1   
POL(false) = 1   
POL(n__f(x1)) = x1   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented: none

(4) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f(X) → if(X, c, n__f(true))
if(true, X, Y) → X
if(false, X, Y) → activate(Y)
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(5) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(6) TRUE