Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 QDPOrderProof (⇔)
        ↳7 QDP
        ↳8 PisEmptyProof (⇔)
        ↳9 TRUE
    ↳10 QDP
        ↳11 QDPOrderProof (⇔)
        ↳12 QDP
        ↳13 PisEmptyProof (⇔)
        ↳14 TRUE

(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) 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:

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

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:

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

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.


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

POL(0) = 1   
POL(SUM(x1, x2)) = x1   
POL(cons(x1, x2)) = 1 + x1 + x2   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(7) 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

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:

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

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.


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: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(WEIGHT(x1)) = x1   
POL(cons(x1, x2)) = 1 + x2   
POL(nil) = 0   
POL(s(x1)) = 0   
POL(sum(x1, x2)) = x2   

The following usable rules [FROCOS05] were oriented:

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

(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

Q is empty.
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