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

(0) Obligation:

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

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, s(x), s(y)) → s(x)
if(false, s(x), s(y)) → s(y)
g(x, c(y)) → c(g(x, y))
g(x, c(y)) → g(x, if(f(x), c(g(s(x), y)), c(y)))

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(s(x)) → F(x)
G(x, c(y)) → G(x, y)
G(x, c(y)) → G(x, if(f(x), c(g(s(x), y)), c(y)))
G(x, c(y)) → IF(f(x), c(g(s(x), y)), c(y))
G(x, c(y)) → F(x)
G(x, c(y)) → G(s(x), y)

The TRS R consists of the following rules:

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, s(x), s(y)) → s(x)
if(false, s(x), s(y)) → s(y)
g(x, c(y)) → c(g(x, y))
g(x, c(y)) → g(x, if(f(x), c(g(s(x), y)), c(y)))

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 2 SCCs with 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(s(x)) → F(x)

The TRS R consists of the following rules:

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, s(x), s(y)) → s(x)
if(false, s(x), s(y)) → s(y)
g(x, c(y)) → c(g(x, y))
g(x, c(y)) → g(x, if(f(x), c(g(s(x), y)), c(y)))

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

(6) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
s(x1)  =  s(x1)

From the DPs we obtained the following set of size-change graphs:

• F(s(x)) → F(x) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(7) TRUE

(8) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(x, c(y)) → G(s(x), y)
G(x, c(y)) → G(x, y)

The TRS R consists of the following rules:

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, s(x), s(y)) → s(x)
if(false, s(x), s(y)) → s(y)
g(x, c(y)) → c(g(x, y))
g(x, c(y)) → g(x, if(f(x), c(g(s(x), y)), c(y)))

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

(9) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
c(x1)  =  c(x1)
s(x1)  =  s

From the DPs we obtained the following set of size-change graphs:

• G(x, c(y)) → G(s(x), y) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 2 > 2

• G(x, c(y)) → G(x, y) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 >= 1, 2 > 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(10) TRUE