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 UsableRulesProof (⇔)
    ↳9 QDP
    ↳10 QReductionProof (⇔)
    ↳11 QDP
    ↳12 UsableRulesReductionPairsProof (⇔)
    ↳13 QDP
    ↳14 MRRProof (⇔)
    ↳15 QDP
    ↳16 PisEmptyProof (⇔)
    ↳17 TRUE
  ↳18 QDP
    ↳19 UsableRulesProof (⇔)
    ↳20 QDP
    ↳21 QReductionProof (⇔)
    ↳22 QDP
    ↳23 QDPOrderProof (⇔)
    ↳24 QDP
    ↳25 DependencyGraphProof (⇔)
    ↳26 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
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) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) 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))

R is empty.
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) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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)

(11) 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))

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

(12) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

SUM(cons(0, x), y) → SUM(x, y)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(SUM(x1, x2)) = x1 + 2·x2   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(s(x1)) = x1   

(13) 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))

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

(14) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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


Used ordering: Polynomial interpretation [POLO]:

POL(SUM(x1, x2)) = 2·x1 + x2   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(s(x1)) = 2 + x1   

(15) Obligation:

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(16) PisEmptyProof (EQUIVALENT transformation)

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

(17) TRUE

(18) 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.

(19) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(20) 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:

empty(nil) → true
empty(cons(n, x)) → false
tail(nil) → nil
tail(cons(n, x)) → x
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

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.

(21) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

head(cons(x0, x1))
weight(x0)
if(true, x0, x1)
if(false, x0, x1)
if2(true, x0)
if2(false, x0)

(22) 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:

empty(nil) → true
empty(cons(n, x)) → false
tail(nil) → nil
tail(cons(n, x)) → x
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

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))

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


IF2(false, x) → WEIGHT(sum(x, cons(0, tail(tail(x)))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(IF2(x1, x2)) = (3/4)x1 + (1/4)x2   
POL(false) = 1/4   
POL(WEIGHT(x1)) = (1/2)x1   
POL(sum(x1, x2)) = (1/4)x1 + x2   
POL(cons(x1, x2)) = 1/4 + (1/4)x1 + (4)x2   
POL(0) = 1/4   
POL(tail(x1)) = (1/4)x1   
POL(IF(x1, x2, x3)) = x2 + (1/4)x3   
POL(empty(x1)) = x1   
POL(s(x1)) = x1   
POL(nil) = 0   
POL(true) = 0   
The value of delta used in the strict ordering is 1/32.
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
empty(nil) → true
empty(cons(n, x)) → false
tail(nil) → nil
tail(cons(n, x)) → x

(24) Obligation:

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

WEIGHT(x) → IF(empty(x), empty(tail(x)), x)
IF(false, b, x) → IF2(b, x)

The TRS R consists of the following rules:

empty(nil) → true
empty(cons(n, x)) → false
tail(nil) → nil
tail(cons(n, x)) → x
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

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))

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

(25) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(26) TRUE