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:

c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(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:

C(c(c(b(x)))) → A(1, b(c(x)))
C(c(c(b(x)))) → B(c(x))
C(c(c(b(x)))) → C(x)
B(c(b(c(x)))) → A(0, a(1, x))
B(c(b(c(x)))) → A(1, x)
A(0, x) → C(c(x))
A(0, x) → C(x)
A(1, x) → C(b(x))
A(1, x) → B(x)

The TRS R consists of the following rules:

c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))

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

(3) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

C(c(c(b(x)))) → B(c(x))
C(c(c(b(x)))) → C(x)
B(c(b(c(x)))) → A(0, a(1, x))
B(c(b(c(x)))) → A(1, x)
A(0, x) → C(x)
A(1, x) → B(x)

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(1) = 0   
POL(A(x₁, x₂)) = 2 + x₁ + x₂   
POL(B(x₁)) = 1 + x₁   
POL(C(x₁)) = x₁   
POL(a(x₁, x₂)) = 4 + x₁ + x₂   
POL(b(x₁)) = 2 + x₁   
POL(c(x₁)) = 2 + x₁   

(4) Obligation:

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

C(c(c(b(x)))) → A(1, b(c(x)))
A(0, x) → C(c(x))
A(1, x) → C(b(x))

The TRS R consists of the following rules:

c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(6) Obligation:

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

A(1, x) → C(b(x))
C(c(c(b(x)))) → A(1, b(c(x)))

The TRS R consists of the following rules:

c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

C(c(c(b(x)))) → A(1, b(c(x)))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(A(x₁, x₂)) = 1 + (3/4)x₂   
POL(1) = 1/2   
POL(C(x₁)) = 1 + (1/2)x₁   
POL(b(x₁)) = (3/2)x₁   
POL(c(x₁)) = 1/4 + (3/2)x₁   
POL(a(x₁, x₂)) = (1/2)x₁ + (9/4)x₂   
POL(0) = 5/4   

The value of delta used in the strict ordering is 1/32.
The following usable rules [FROCOS05] were oriented:

a(1, x) → c(b(x))
c(c(c(b(x)))) → a(1, b(c(x)))
a(0, x) → c(c(x))
b(c(b(c(x)))) → a(0, a(1, x))

(8) Obligation:

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

A(1, x) → C(b(x))

The TRS R consists of the following rules:

c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))

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

(9) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(10) TRUE