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
empty(nil) → true
empty(cons(n, x)) → false
tail(nil) → nil
tail(cons(n, x)) → x
head(cons(n, x)) → n
weight(x) → if(empty(x), empty(tail(x)), x)
if(true, b, x) → weight_undefined_error
if(false, b, x) → if2(b, x)
if2(true, x) → head(x)
if2(false, x) → weight(sum(x, cons(0, tail(tail(x)))))

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
empty(nil) → true
empty(cons(n, x)) → false
tail(nil) → nil
tail(cons(n, x)) → x
head(cons(n, x)) → n
weight(x) → if(empty(x), empty(tail(x)), x)
if(true, b, x) → weight_undefined_error
if(false, b, x) → if2(b, x)
if2(true, x) → head(x)
if2(false, x) → weight(sum(x, cons(0, tail(tail(x)))))

The set Q consists of the following terms:

sum(cons(s(x0), x1), cons(x2, x3))
sum(cons(0, x0), x1)
sum(nil, x0)
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
weight(x0)
if(true, x0, x1)
if(false, x0, x1)
if2(true, x0)
if2(false, x0)

(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(x) → IF(empty(x), empty(tail(x)), x)
WEIGHT(x) → EMPTY(x)
WEIGHT(x) → EMPTY(tail(x))
WEIGHT(x) → TAIL(x)
IF(false, b, x) → IF2(b, x)
IF2(true, x) → HEAD(x)
IF2(false, x) → WEIGHT(sum(x, cons(0, tail(tail(x)))))
IF2(false, x) → SUM(x, cons(0, tail(tail(x))))
IF2(false, x) → TAIL(tail(x))
IF2(false, x) → TAIL(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
empty(nil) → true
empty(cons(n, x)) → false
tail(nil) → nil
tail(cons(n, x)) → x
head(cons(n, x)) → n
weight(x) → if(empty(x), empty(tail(x)), x)
if(true, b, x) → weight_undefined_error
if(false, b, x) → if2(b, x)
if2(true, x) → head(x)
if2(false, x) → weight(sum(x, cons(0, tail(tail(x)))))

The set Q consists of the following terms:

sum(cons(s(x0), x1), cons(x2, x3))
sum(cons(0, x0), x1)
sum(nil, x0)
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
weight(x0)
if(true, x0, x1)
if(false, x0, x1)
if2(true, x0)
if2(false, x0)

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 7 less nodes.

(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
empty(nil) → true
empty(cons(n, x)) → false
tail(nil) → nil
tail(cons(n, x)) → x
head(cons(n, x)) → n
weight(x) → if(empty(x), empty(tail(x)), x)
if(true, b, x) → weight_undefined_error
if(false, b, x) → if2(b, x)
if2(true, x) → head(x)
if2(false, x) → weight(sum(x, cons(0, tail(tail(x)))))

The set Q consists of the following terms:

sum(cons(s(x0), x1), cons(x2, x3))
sum(cons(0, x0), x1)
sum(nil, x0)
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
weight(x0)
if(true, x0, x1)
if(false, x0, x1)
if2(true, x0)
if2(false, x0)

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(x2)
0  =  0
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

SUM₂ > s₁
cons₁ > s₁
0 > s₁

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
empty(nil) → true
empty(cons(n, x)) → false
tail(nil) → nil
tail(cons(n, x)) → x
head(cons(n, x)) → n
weight(x) → if(empty(x), empty(tail(x)), x)
if(true, b, x) → weight_undefined_error
if(false, b, x) → if2(b, x)
if2(true, x) → head(x)
if2(false, x) → weight(sum(x, cons(0, tail(tail(x)))))

The set Q consists of the following terms:

sum(cons(s(x0), x1), cons(x2, x3))
sum(cons(0, x0), x1)
sum(nil, x0)
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
weight(x0)
if(true, x0, x1)
if(false, x0, x1)
if2(true, x0)
if2(false, x0)

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:

s₁ > cons₂ > 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
empty(nil) → true
empty(cons(n, x)) → false
tail(nil) → nil
tail(cons(n, x)) → x
head(cons(n, x)) → n
weight(x) → if(empty(x), empty(tail(x)), x)
if(true, b, x) → weight_undefined_error
if(false, b, x) → if2(b, x)
if2(true, x) → head(x)
if2(false, x) → weight(sum(x, cons(0, tail(tail(x)))))

The set Q consists of the following terms:

sum(cons(s(x0), x1), cons(x2, x3))
sum(cons(0, x0), x1)
sum(nil, x0)
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
weight(x0)
if(true, x0, x1)
if(false, x0, x1)
if2(true, x0)
if2(false, x0)

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:

IF2(false, x) → WEIGHT(sum(x, cons(0, tail(tail(x)))))
WEIGHT(x) → IF(empty(x), empty(tail(x)), x)
IF(false, b, x) → IF2(b, 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
empty(nil) → true
empty(cons(n, x)) → false
tail(nil) → nil
tail(cons(n, x)) → x
head(cons(n, x)) → n
weight(x) → if(empty(x), empty(tail(x)), x)
if(true, b, x) → weight_undefined_error
if(false, b, x) → if2(b, x)
if2(true, x) → head(x)
if2(false, x) → weight(sum(x, cons(0, tail(tail(x)))))

The set Q consists of the following terms:

sum(cons(s(x0), x1), cons(x2, x3))
sum(cons(0, x0), x1)
sum(nil, x0)
empty(nil)
empty(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
head(cons(x0, x1))
weight(x0)
if(true, x0, x1)
if(false, x0, x1)
if2(true, x0)
if2(false, x0)

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