Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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

sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) → sum(x, y)
sum(nil, y) → y
weight(cons(n, cons(m, x))) → weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) → n

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:

sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) → sum(x, y)
sum(nil, y) → y
weight(cons(n, cons(m, x))) → weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) → n

The set Q consists of the following terms:

sum(cons(s(x0), x1), cons(x2, x3))
sum(cons(0, x0), x1)
sum(nil, x0)
weight(cons(x0, cons(x1, x2)))
weight(cons(x0, nil))

(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:

SUM(cons(s(n), x), cons(m, y)) → SUM(cons(n, x), cons(s(m), y))
SUM(cons(0, x), y) → SUM(x, y)
WEIGHT(cons(n, cons(m, x))) → WEIGHT(sum(cons(n, cons(m, x)), cons(0, x)))
WEIGHT(cons(n, cons(m, x))) → SUM(cons(n, cons(m, x)), cons(0, x))

The TRS R consists of the following rules:

sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) → sum(x, y)
sum(nil, y) → y
weight(cons(n, cons(m, x))) → weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) → n

The set Q consists of the following terms:

sum(cons(s(x0), x1), cons(x2, x3))
sum(cons(0, x0), x1)
sum(nil, x0)
weight(cons(x0, cons(x1, x2)))
weight(cons(x0, nil))

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 with 1 less node.

(6) Complex Obligation (AND)

(7) Obligation:

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

SUM(cons(0, x), y) → SUM(x, y)
SUM(cons(s(n), x), cons(m, y)) → SUM(cons(n, x), cons(s(m), y))

The TRS R consists of the following rules:

sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) → sum(x, y)
sum(nil, y) → y
weight(cons(n, cons(m, x))) → weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) → n

The set Q consists of the following terms:

sum(cons(s(x0), x1), cons(x2, x3))
sum(cons(0, x0), x1)
sum(nil, x0)
weight(cons(x0, cons(x1, x2)))
weight(cons(x0, nil))

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.

SUM(cons(0, x), y) → SUM(x, y)

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

Recursive Path Order [RPO].
Precedence:

weight₁ > cons₂ > SUM₁
weight₁ > cons₂ > 0

The following usable rules [FROCOS05] were oriented:

sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) → sum(x, y)
sum(nil, y) → y
weight(cons(n, cons(m, x))) → weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) → n

(9) Obligation:

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

SUM(cons(s(n), x), cons(m, y)) → SUM(cons(n, x), cons(s(m), y))

The TRS R consists of the following rules:

sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) → sum(x, y)
sum(nil, y) → y
weight(cons(n, cons(m, x))) → weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) → n

The set Q consists of the following terms:

sum(cons(s(x0), x1), cons(x2, x3))
sum(cons(0, x0), x1)
sum(nil, x0)
weight(cons(x0, cons(x1, x2)))
weight(cons(x0, nil))

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

(10) Obligation:

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

WEIGHT(cons(n, cons(m, x))) → WEIGHT(sum(cons(n, cons(m, x)), cons(0, x)))

The TRS R consists of the following rules:

sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) → sum(x, y)
sum(nil, y) → y
weight(cons(n, cons(m, x))) → weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) → n

The set Q consists of the following terms:

sum(cons(s(x0), x1), cons(x2, x3))
sum(cons(0, x0), x1)
sum(nil, x0)
weight(cons(x0, cons(x1, x2)))
weight(cons(x0, nil))

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.

WEIGHT(cons(n, cons(m, x))) → WEIGHT(sum(cons(n, cons(m, x)), cons(0, x)))

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

Recursive Path Order [RPO].
Precedence:

cons₂ > WEIGHT₁ > sum₁ > 0
cons₂ > weight₁ > sum₁ > 0
nil > 0

The following usable rules [FROCOS05] were oriented:

sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) → sum(x, y)
sum(nil, y) → y
weight(cons(n, cons(m, x))) → weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) → n

(12) Obligation:

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

sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) → sum(x, y)
sum(nil, y) → y
weight(cons(n, cons(m, x))) → weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) → n

The set Q consists of the following terms:

sum(cons(s(x0), x1), cons(x2, x3))
sum(cons(0, x0), x1)
sum(nil, x0)
weight(cons(x0, cons(x1, x2)))
weight(cons(x0, nil))

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