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

(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) 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:

LT(s(x), s(y)) → LT(x, y)

R is empty.
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) 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].

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)

(11) Obligation:

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

LT(s(x), s(y)) → LT(x, y)

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• LT(s(x), s(y)) → LT(x, y)
  The graph contains the following edges 1 > 1, 2 > 2

(13) TRUE

(14) 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.

(15) 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.

(16) Obligation:

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

LENGTH(cons(x, l)) → LENGTH(l)

R is empty.
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.

(17) 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].

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)

(18) Obligation:

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

LENGTH(cons(x, l)) → LENGTH(l)

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

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• LENGTH(cons(x, l)) → LENGTH(l)
  The graph contains the following edges 1 > 1

(20) TRUE

(21) 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.

(22) 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.

(23) 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:

tail(nil) → nil
tail(cons(x, l)) → l
head(cons(x, l)) → x
head(nil) → undefined
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)

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.

(24) 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].

reverse(x0)
rev(x0, x1, x2, x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)

(25) 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:

tail(nil) → nil
tail(cons(x, l)) → l
head(cons(x, l)) → x
head(nil) → undefined
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)

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))

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

(26) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule REV(x, l, accu, orig) → IF(lt(x, length(orig)), x, l, accu, orig) we obtained the following new rules [LPAR04]:

REV(s(z0), y_0, cons(y_1, z2), z3) → IF(lt(s(z0), length(z3)), s(z0), y_0, cons(y_1, z2), z3)

(27) Obligation:

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

IF(true, x, l, accu, orig) → REV(s(x), tail(l), cons(head(l), accu), orig)
REV(s(z0), y_0, cons(y_1, z2), z3) → IF(lt(s(z0), length(z3)), s(z0), y_0, cons(y_1, z2), z3)

The TRS R consists of the following rules:

tail(nil) → nil
tail(cons(x, l)) → l
head(cons(x, l)) → x
head(nil) → undefined
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)

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))

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

(28) NonInfProof (EQUIVALENT transformation)

The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
Note that final constraints are written in bold face.


For Pair REV(x, l, accu, orig) → IF(lt(x, length(orig)), x, l, accu, orig) the following chains were created:

• We consider the chain IF(true, x, l, accu, orig) → REV(s(x), tail(l), cons(head(l), accu), orig), REV(x, l, accu, orig) → IF(lt(x, length(orig)), x, l, accu, orig) which results in the following constraint:
  

  (1)    (REV(s(x4), tail(x5), cons(head(x5), x6), x7)=REV(x8, x9, x10, x11) ⇒ REV(x8, x9, x10, x11)≥IF(lt(x8, length(x11)), x8, x9, x10, x11))

  
  
  We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint:
  

  (2)    (REV(s(x4), x9, x10, x7)≥IF(lt(s(x4), length(x7)), s(x4), x9, x10, x7))

  
  
  

For Pair IF(true, x, l, accu, orig) → REV(s(x), tail(l), cons(head(l), accu), orig) the following chains were created:

• We consider the chain 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) which results in the following constraint:
  

  (3)    (IF(lt(x12, length(x15)), x12, x13, x14, x15)=IF(true, x16, x17, x18, x19) ⇒ IF(true, x16, x17, x18, x19)≥REV(s(x16), tail(x17), cons(head(x17), x18), x19))

  
  
  We simplified constraint (3) using rules (I), (II), (III), (VII) which results in the following new constraint:
  

  (4)    (length(x15)=x24∧lt(x12, x24)=true ⇒ IF(true, x12, x13, x14, x15)≥REV(s(x12), tail(x13), cons(head(x13), x14), x15))

  
  
  We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on lt(x12, x24)=true which results in the following new constraints:
  

  (5)    (true=true∧length(x15)=s(x26) ⇒ IF(true, 0, x13, x14, x15)≥REV(s(0), tail(x13), cons(head(x13), x14), x15))

  
  

  (6)    (lt(x28, x27)=true∧length(x15)=s(x27)∧(∀x29,x30,x31:lt(x28, x27)=true∧length(x29)=x27 ⇒ IF(true, x28, x30, x31, x29)≥REV(s(x28), tail(x30), cons(head(x30), x31), x29)) ⇒ IF(true, s(x28), x13, x14, x15)≥REV(s(s(x28)), tail(x13), cons(head(x13), x14), x15))

  
  
  We simplified constraint (5) using rules (I), (II) which results in the following new constraint:
  

  (7)    (length(x15)=s(x26) ⇒ IF(true, 0, x13, x14, x15)≥REV(s(0), tail(x13), cons(head(x13), x14), x15))

  
  
  We simplified constraint (6) using rule (V) (with possible (I) afterwards) using induction on length(x15)=s(x27) which results in the following new constraint:
  

  (8)    (s(length(x37))=s(x27)∧lt(x28, x27)=true∧(∀x29,x30,x31:lt(x28, x27)=true∧length(x29)=x27 ⇒ IF(true, x28, x30, x31, x29)≥REV(s(x28), tail(x30), cons(head(x30), x31), x29))∧(∀x39,x40,x41,x42,x43,x44,x45:length(x37)=s(x39)∧lt(x40, x39)=true∧(∀x41,x42,x43:lt(x40, x39)=true∧length(x41)=x39 ⇒ IF(true, x40, x42, x43, x41)≥REV(s(x40), tail(x42), cons(head(x42), x43), x41)) ⇒ IF(true, s(x40), x44, x45, x37)≥REV(s(s(x40)), tail(x44), cons(head(x44), x45), x37)) ⇒ IF(true, s(x28), x13, x14, cons(x38, x37))≥REV(s(s(x28)), tail(x13), cons(head(x13), x14), cons(x38, x37)))

  
  
  We simplified constraint (7) using rule (V) (with possible (I) afterwards) using induction on length(x15)=s(x26) which results in the following new constraint:
  

  (9)    (s(length(x32))=s(x26)∧(∀x34,x35,x36:length(x32)=s(x34) ⇒ IF(true, 0, x35, x36, x32)≥REV(s(0), tail(x35), cons(head(x35), x36), x32)) ⇒ IF(true, 0, x13, x14, cons(x33, x32))≥REV(s(0), tail(x13), cons(head(x13), x14), cons(x33, x32)))

  
  
  We simplified constraint (9) using rules (I), (II), (IV) which results in the following new constraint:
  

  (10)    (IF(true, 0, x13, x14, cons(x33, x32))≥REV(s(0), tail(x13), cons(head(x13), x14), cons(x33, x32)))

  
  
  We simplified constraint (8) using rules (I), (II) which results in the following new constraint:
  

  (11)    (length(x37)=x27∧lt(x28, x27)=true∧(∀x29,x30,x31:lt(x28, x27)=true∧length(x29)=x27 ⇒ IF(true, x28, x30, x31, x29)≥REV(s(x28), tail(x30), cons(head(x30), x31), x29))∧(∀x39,x40,x41,x42,x43,x44,x45:length(x37)=s(x39)∧lt(x40, x39)=true∧(∀x41,x42,x43:lt(x40, x39)=true∧length(x41)=x39 ⇒ IF(true, x40, x42, x43, x41)≥REV(s(x40), tail(x42), cons(head(x42), x43), x41)) ⇒ IF(true, s(x40), x44, x45, x37)≥REV(s(s(x40)), tail(x44), cons(head(x44), x45), x37)) ⇒ IF(true, s(x28), x13, x14, cons(x38, x37))≥REV(s(s(x28)), tail(x13), cons(head(x13), x14), cons(x38, x37)))

  
  
  We simplified constraint (11) using rule (VI) where we applied the induction hypothesis (∀x29,x30,x31:lt(x28, x27)=true∧length(x29)=x27 ⇒ IF(true, x28, x30, x31, x29)≥REV(s(x28), tail(x30), cons(head(x30), x31), x29)) with σ = [x29 / x37, x31 / x14, x30 / x13] which results in the following new constraint:
  

  (12)    (IF(true, x28, x13, x14, x37)≥REV(s(x28), tail(x13), cons(head(x13), x14), x37)∧(∀x39,x40,x41,x42,x43,x44,x45:length(x37)=s(x39)∧lt(x40, x39)=true∧(∀x41,x42,x43:lt(x40, x39)=true∧length(x41)=x39 ⇒ IF(true, x40, x42, x43, x41)≥REV(s(x40), tail(x42), cons(head(x42), x43), x41)) ⇒ IF(true, s(x40), x44, x45, x37)≥REV(s(s(x40)), tail(x44), cons(head(x44), x45), x37)) ⇒ IF(true, s(x28), x13, x14, cons(x38, x37))≥REV(s(s(x28)), tail(x13), cons(head(x13), x14), cons(x38, x37)))

  
  
  We simplified constraint (12) using rule (IV) which results in the following new constraint:
  

  (13)    (IF(true, x28, x13, x14, x37)≥REV(s(x28), tail(x13), cons(head(x13), x14), x37) ⇒ IF(true, s(x28), x13, x14, cons(x38, x37))≥REV(s(s(x28)), tail(x13), cons(head(x13), x14), cons(x38, x37)))

  
  
  

To summarize, we get the following constraints P_≥ for the following pairs.

• REV(x, l, accu, orig) → IF(lt(x, length(orig)), x, l, accu, orig)
  
  • (REV(s(x4), x9, x10, x7)≥IF(lt(s(x4), length(x7)), s(x4), x9, x10, x7))
    
  
  
• IF(true, x, l, accu, orig) → REV(s(x), tail(l), cons(head(l), accu), orig)
  
  • (IF(true, 0, x13, x14, cons(x33, x32))≥REV(s(0), tail(x13), cons(head(x13), x14), cons(x33, x32)))
    
  • (IF(true, x28, x13, x14, x37)≥REV(s(x28), tail(x13), cons(head(x13), x14), x37) ⇒ IF(true, s(x28), x13, x14, cons(x38, x37))≥REV(s(s(x28)), tail(x13), cons(head(x13), x14), cons(x38, x37)))
    
  
  

The constraints for P_> respective P_bound are constructed from P_≥ where we just replace every occurence of "t ≥ s" in P_≥ by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for P_bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [NONINF]:

POL(0) = 1   
POL(IF(x₁, x₂, x₃, x₄, x₅)) = -x₁ - x₂ + x₅   
POL(REV(x₁, x₂, x₃, x₄)) = -x₁ + x₄   
POL(c) = -1   
POL(cons(x₁, x₂)) = 1 + x₂   
POL(false) = 0   
POL(head(x₁)) = x₁   
POL(length(x₁)) = 0   
POL(lt(x₁, x₂)) = 0   
POL(nil) = 1   
POL(s(x₁)) = 1 + x₁   
POL(tail(x₁)) = 0   
POL(true) = 0   
POL(undefined) = 1   

The following pairs are in P_>:

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

The following pairs are in P_bound:

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

The following rules are usable:

true → lt(0, s(y))
lt(x, y) → lt(s(x), s(y))
false → lt(x, 0)

(29) 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)

The TRS R consists of the following rules:

tail(nil) → nil
tail(cons(x, l)) → l
head(cons(x, l)) → x
head(nil) → undefined
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)

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))

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

(30) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(31) TRUE