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:

int(0, 0) → .(0, nil)
int(0, s(y)) → .(0, int(s(0), s(y)))
int(s(x), 0) → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(x, y)) → .(s(x), int_list(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:

INT(0, s(y)) → INT(s(0), s(y))
INT(s(x), s(y)) → INT_LIST(int(x, y))
INT(s(x), s(y)) → INT(x, y)
INT_LIST(.(x, y)) → INT_LIST(y)

The TRS R consists of the following rules:

int(0, 0) → .(0, nil)
int(0, s(y)) → .(0, int(s(0), s(y)))
int(s(x), 0) → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(x, y)) → .(s(x), int_list(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 1 less node.

(4) Complex Obligation (AND)

(5) Obligation:

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

INT_LIST(.(x, y)) → INT_LIST(y)

The TRS R consists of the following rules:

int(0, 0) → .(0, nil)
int(0, s(y)) → .(0, int(s(0), s(y)))
int(s(x), 0) → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(x, y)) → .(s(x), int_list(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.

INT_LIST(.(x, y)) → INT_LIST(y)

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

Recursive path order with status [RPO].
Quasi-Precedence:

trivial

Status:

.₂: multiset

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

int(0, 0) → .(0, nil)
int(0, s(y)) → .(0, int(s(0), s(y)))
int(s(x), 0) → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(x, y)) → .(s(x), int_list(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:

INT(s(x), s(y)) → INT(x, y)
INT(0, s(y)) → INT(s(0), s(y))

The TRS R consists of the following rules:

int(0, 0) → .(0, nil)
int(0, s(y)) → .(0, int(s(0), s(y)))
int(s(x), 0) → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(x, y)) → .(s(x), int_list(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.

INT(s(x), s(y)) → INT(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
INT(x1, x2)  =  x2
s(x1)  =  s(x1)
0  =  0

Recursive path order with status [RPO].
Quasi-Precedence:

[s₁, 0]

Status:

s₁: [1]
0: multiset

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

INT(0, s(y)) → INT(s(0), s(y))

The TRS R consists of the following rules:

int(0, 0) → .(0, nil)
int(0, s(y)) → .(0, int(s(0), s(y)))
int(s(x), 0) → nil
int(s(x), s(y)) → int_list(int(x, y))
int_list(nil) → nil
int_list(.(x, y)) → .(s(x), int_list(y))

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

(13) DependencyGraphProof (EQUIVALENT transformation)

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

(14) TRUE