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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 QDPOrderProof (⇔)
    ↳11 QDP
    ↳12 PisEmptyProof (⇔)
    ↳13 TRUE
  ↳14 QDP
    ↳15 QDPOrderProof (⇔)
    ↳16 QDP
    ↳17 PisEmptyProof (⇔)
    ↳18 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) 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, x2)
cons(x1, x2)  =  cons(x1, x2)
0  =  0
s(x1)  =  x1

Recursive Path Order [RPO].
Precedence:
cons2 > [SUM2, 0]


The following usable rules [FROCOS05] were oriented: none

(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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: Combined order from the following AFS and order.
SUM(x1, x2)  =  SUM(x1)
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:
[cons2, s1] > SUM1


The following usable rules [FROCOS05] were oriented: none

(11) 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.

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) TRUE

(14) 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.

(15) 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)  =  x1
cons(x1, x2)  =  cons(x1, x2)
sum(x1, x2)  =  sum(x2)
0  =  0
nil  =  nil
s(x1)  =  s

Recursive Path Order [RPO].
Precedence:
cons2 > 0 > [sum1, s]
nil > [sum1, s]


The following usable rules [FROCOS05] were oriented:

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

(16) 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.

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE