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

The TRS R consists of the following rules:

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), 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 2 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, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), 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)  =  x1
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

s₁: [1]

The following usable rules [FROCOS05] were oriented: none

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

The TRS R consists of the following rules:

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), 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(c(y)), y)
G(x, c(y)) → G(x, g(s(c(y)), 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
g(x1, x2)  =  g(x1, x2)
if(x1, x2, x3)  =  if(x1, x2, x3)
f(x1)  =  f
0  =  0
true  =  true
1  =  1
false  =  false

Lexicographic path order with status [LPO].
Quasi-Precedence:

[G₂, c₁] > s > [f, true] > false > [g₂, if₃]
0 > [g₂, if₃]
1 > false > [g₂, if₃]

Status:

G₂: [2,1]
c₁: [1]
s: []
g₂: [1,2]
if₃: [3,1,2]
f: []
0: []
true: []
1: []
false: []

The following usable rules [FROCOS05] were oriented:

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