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(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))

The set Q consists of the following terms:

f(nil)
f(.(nil, x0))
f(.(.(x0, x1), x2))
g(nil)
g(.(x0, nil))
g(.(x0, .(x1, x2)))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

F(.(nil, y)) → F(y)
F(.(.(x, y), z)) → F(.(x, .(y, z)))
G(.(x, nil)) → G(x)
G(.(x, .(y, z))) → G(.(.(x, y), z))

The TRS R consists of the following rules:

f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))

The set Q consists of the following terms:

f(nil)
f(.(nil, x0))
f(.(.(x0, x1), x2))
g(nil)
g(.(x0, nil))
g(.(x0, .(x1, x2)))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.

(6) Complex Obligation (AND)

(7) Obligation:

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

G(.(x, .(y, z))) → G(.(.(x, y), z))
G(.(x, nil)) → G(x)

The TRS R consists of the following rules:

f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))

The set Q consists of the following terms:

f(nil)
f(.(nil, x0))
f(.(.(x0, x1), x2))
g(nil)
g(.(x0, nil))
g(.(x0, .(x1, x2)))

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

G(.(x, .(y, z))) → G(.(.(x, y), z))
G(.(x, nil)) → G(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
G(x1)  =  x1
.(x1, x2)  =  .(x1, x2)
nil  =  nil

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

trivial

Status:

.₂: [2,1]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))

The set Q consists of the following terms:

f(nil)
f(.(nil, x0))
f(.(.(x0, x1), x2))
g(nil)
g(.(x0, nil))
g(.(x0, .(x1, x2)))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

F(.(.(x, y), z)) → F(.(x, .(y, z)))
F(.(nil, y)) → F(y)

The TRS R consists of the following rules:

f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))

The set Q consists of the following terms:

f(nil)
f(.(nil, x0))
f(.(.(x0, x1), x2))
g(nil)
g(.(x0, nil))
g(.(x0, .(x1, x2)))

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F(.(.(x, y), z)) → F(.(x, .(y, z)))
F(.(nil, y)) → F(y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1)  =  x1
.(x1, x2)  =  .(x1, x2)
nil  =  nil

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

trivial

Status:

.₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))

The set Q consists of the following terms:

f(nil)
f(.(nil, x0))
f(.(.(x0, x1), x2))
g(nil)
g(.(x0, nil))
g(.(x0, .(x1, x2)))

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE