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:

length(nil) → 0
length(cons(x, l)) → s(length(l))
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
head(cons(x, l)) → x
head(nil) → undefined
tail(nil) → nil
tail(cons(x, l)) → l
reverse(l) → rev(0, l, nil, l)
rev(x, l, accu, orig) → if(lt(x, length(orig)), x, l, accu, orig)
if(true, x, l, accu, orig) → rev(s(x), tail(l), cons(head(l), accu), orig)
if(false, x, l, accu, orig) → accu

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:

length(nil) → 0
length(cons(x, l)) → s(length(l))
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
head(cons(x, l)) → x
head(nil) → undefined
tail(nil) → nil
tail(cons(x, l)) → l
reverse(l) → rev(0, l, nil, l)
rev(x, l, accu, orig) → if(lt(x, length(orig)), x, l, accu, orig)
if(true, x, l, accu, orig) → rev(s(x), tail(l), cons(head(l), accu), orig)
if(false, x, l, accu, orig) → accu

The set Q consists of the following terms:

length(nil)
length(cons(x0, x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
head(cons(x0, x1))
head(nil)
tail(nil)
tail(cons(x0, x1))
reverse(x0)
rev(x0, x1, x2, x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

(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:

LENGTH(cons(x, l)) → LENGTH(l)
LT(s(x), s(y)) → LT(x, y)
REVERSE(l) → REV(0, l, nil, l)
REV(x, l, accu, orig) → IF(lt(x, length(orig)), x, l, accu, orig)
REV(x, l, accu, orig) → LT(x, length(orig))
REV(x, l, accu, orig) → LENGTH(orig)
IF(true, x, l, accu, orig) → REV(s(x), tail(l), cons(head(l), accu), orig)
IF(true, x, l, accu, orig) → TAIL(l)
IF(true, x, l, accu, orig) → HEAD(l)

The TRS R consists of the following rules:

length(nil) → 0
length(cons(x, l)) → s(length(l))
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
head(cons(x, l)) → x
head(nil) → undefined
tail(nil) → nil
tail(cons(x, l)) → l
reverse(l) → rev(0, l, nil, l)
rev(x, l, accu, orig) → if(lt(x, length(orig)), x, l, accu, orig)
if(true, x, l, accu, orig) → rev(s(x), tail(l), cons(head(l), accu), orig)
if(false, x, l, accu, orig) → accu

The set Q consists of the following terms:

length(nil)
length(cons(x0, x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
head(cons(x0, x1))
head(nil)
tail(nil)
tail(cons(x0, x1))
reverse(x0)
rev(x0, x1, x2, x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

length(nil) → 0
length(cons(x, l)) → s(length(l))
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
head(cons(x, l)) → x
head(nil) → undefined
tail(nil) → nil
tail(cons(x, l)) → l
reverse(l) → rev(0, l, nil, l)
rev(x, l, accu, orig) → if(lt(x, length(orig)), x, l, accu, orig)
if(true, x, l, accu, orig) → rev(s(x), tail(l), cons(head(l), accu), orig)
if(false, x, l, accu, orig) → accu

The set Q consists of the following terms:

length(nil)
length(cons(x0, x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
head(cons(x0, x1))
head(nil)
tail(nil)
tail(cons(x0, x1))
reverse(x0)
rev(x0, x1, x2, x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

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.

LT(s(x), s(y)) → LT(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LT(x1, x2)  =  LT(x2)
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

s₁ > LT₁

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

length(nil) → 0
length(cons(x, l)) → s(length(l))
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
head(cons(x, l)) → x
head(nil) → undefined
tail(nil) → nil
tail(cons(x, l)) → l
reverse(l) → rev(0, l, nil, l)
rev(x, l, accu, orig) → if(lt(x, length(orig)), x, l, accu, orig)
if(true, x, l, accu, orig) → rev(s(x), tail(l), cons(head(l), accu), orig)
if(false, x, l, accu, orig) → accu

The set Q consists of the following terms:

length(nil)
length(cons(x0, x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
head(cons(x0, x1))
head(nil)
tail(nil)
tail(cons(x0, x1))
reverse(x0)
rev(x0, x1, x2, x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

LENGTH(cons(x, l)) → LENGTH(l)

The TRS R consists of the following rules:

length(nil) → 0
length(cons(x, l)) → s(length(l))
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
head(cons(x, l)) → x
head(nil) → undefined
tail(nil) → nil
tail(cons(x, l)) → l
reverse(l) → rev(0, l, nil, l)
rev(x, l, accu, orig) → if(lt(x, length(orig)), x, l, accu, orig)
if(true, x, l, accu, orig) → rev(s(x), tail(l), cons(head(l), accu), orig)
if(false, x, l, accu, orig) → accu

The set Q consists of the following terms:

length(nil)
length(cons(x0, x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
head(cons(x0, x1))
head(nil)
tail(nil)
tail(cons(x0, x1))
reverse(x0)
rev(x0, x1, x2, x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LENGTH(cons(x, l)) → LENGTH(l)

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive Path Order [RPO].
Precedence:

cons₂ > LENGTH₁

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

length(nil) → 0
length(cons(x, l)) → s(length(l))
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
head(cons(x, l)) → x
head(nil) → undefined
tail(nil) → nil
tail(cons(x, l)) → l
reverse(l) → rev(0, l, nil, l)
rev(x, l, accu, orig) → if(lt(x, length(orig)), x, l, accu, orig)
if(true, x, l, accu, orig) → rev(s(x), tail(l), cons(head(l), accu), orig)
if(false, x, l, accu, orig) → accu

The set Q consists of the following terms:

length(nil)
length(cons(x0, x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
head(cons(x0, x1))
head(nil)
tail(nil)
tail(cons(x0, x1))
reverse(x0)
rev(x0, x1, x2, x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

REV(x, l, accu, orig) → IF(lt(x, length(orig)), x, l, accu, orig)
IF(true, x, l, accu, orig) → REV(s(x), tail(l), cons(head(l), accu), orig)

The TRS R consists of the following rules:

length(nil) → 0
length(cons(x, l)) → s(length(l))
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
head(cons(x, l)) → x
head(nil) → undefined
tail(nil) → nil
tail(cons(x, l)) → l
reverse(l) → rev(0, l, nil, l)
rev(x, l, accu, orig) → if(lt(x, length(orig)), x, l, accu, orig)
if(true, x, l, accu, orig) → rev(s(x), tail(l), cons(head(l), accu), orig)
if(false, x, l, accu, orig) → accu

The set Q consists of the following terms:

length(nil)
length(cons(x0, x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
head(cons(x0, x1))
head(nil)
tail(nil)
tail(cons(x0, x1))
reverse(x0)
rev(x0, x1, x2, x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

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