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:

intlist(nil) → nil
int(s(x), 0) → nil
int(x, x) → cons(x, nil)
intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))
intlist(cons(x, nil)) → cons(s(x), nil)

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:

INTLIST(cons(x, y)) → INTLIST(y)
INT(s(x), s(y)) → INTLIST(int(x, y))
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:

intlist(nil) → nil
int(s(x), 0) → nil
int(x, x) → cons(x, nil)
intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))
intlist(cons(x, nil)) → cons(s(x), nil)

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:

INTLIST(cons(x, y)) → INTLIST(y)

The TRS R consists of the following rules:

intlist(nil) → nil
int(s(x), 0) → nil
int(x, x) → cons(x, nil)
intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))
intlist(cons(x, nil)) → cons(s(x), nil)

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.

INTLIST(cons(x, y)) → INTLIST(y)

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
INTLIST(x0, x1)  =  INTLIST(x1)

Tags:
INTLIST has argument tags [0,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
INTLIST(x1)  =  INTLIST
cons(x1, x2)  =  cons(x1, x2)

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

trivial

Status:

INTLIST: []
cons₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

intlist(nil) → nil
int(s(x), 0) → nil
int(x, x) → cons(x, nil)
intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))
intlist(cons(x, nil)) → cons(s(x), nil)

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

The TRS R consists of the following rules:

intlist(nil) → nil
int(s(x), 0) → nil
int(x, x) → cons(x, nil)
intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))
intlist(cons(x, nil)) → cons(s(x), nil)

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: SCNP Order with the following components:
Level mapping:
Top level AFS:
INT(x0, x1, x2)  =  INT(x2)

Tags:
INT has argument tags [1,1,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
INT(x1, x2)  =  x2
0  =  0
s(x1)  =  s(x1)

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

0 > s₁

Status:

0: []
s₁: [1]

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:

intlist(nil) → nil
int(s(x), 0) → nil
int(x, x) → cons(x, nil)
intlist(cons(x, y)) → cons(s(x), intlist(y))
int(s(x), s(y)) → intlist(int(x, y))
int(0, s(y)) → cons(0, int(s(0), s(y)))
intlist(cons(x, nil)) → cons(s(x), nil)

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