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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F(s(x)) → F(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1)  =  F(x1)
s(x1)  =  s(x1)
f(x1)  =  f(x1)
0  =  0
true  =  true
1  =  1
false  =  false
if(x1, x2, x3)  =  if(x1, x2, x3)
g(x1, x2)  =  g
c(x1)  =  c

Recursive path order with status [RPO].
Precedence:

F₁ > c
0 > c
1 > c
g > f₁ > true > c
g > f₁ > false > s₁ > c
g > if₃ > s₁ > c

Status:

F₁: [1]
s₁: multiset
f₁: multiset
0: multiset
true: multiset
1: multiset
false: multiset
if₃: multiset
g: multiset
c: multiset

The following usable rules [FROCOS05] were oriented:

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)))

(7) Obligation:

Q DP problem:
P is empty.
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.

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) 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.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
G(x1, x2)  =  G(x1, x2)
c(x1)  =  c(x1)
s(x1)  =  s
f(x1)  =  f
0  =  0
true  =  true
1  =  1
false  =  false
if(x1, x2, x3)  =  if
g(x1, x2)  =  g(x1, x2)

Recursive path order with status [RPO].
Precedence:

f > true > s
f > false > s
0 > true > s
1 > s
g₂ > c₁ > G₂ > s
g₂ > c₁ > if > s

Status:

G₂: multiset
c₁: multiset
s: multiset
f: []
0: multiset
true: multiset
1: multiset
false: multiset
if: multiset
g₂: [2,1]

The following usable rules [FROCOS05] were oriented:

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)))

(12) Obligation:

Q DP problem:
P is empty.
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.

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE