Termination w.r.t. Q proof of /tmp/aprove_benchmark_example

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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.

↳ QTRS

  ↳ Overlay + Local Confluence

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.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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)

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

IF(true, x, l, accu, orig) → HEAD(l)
REV(x, l, accu, orig) → IF(lt(x, length(orig)), x, l, accu, orig)
LT(s(x), s(y)) → LT(x, y)
REV(x, l, accu, orig) → LT(x, length(orig))
LENGTH(cons(x, l)) → LENGTH(l)
IF(true, x, l, accu, orig) → REV(s(x), tail(l), cons(head(l), accu), orig)
IF(true, x, l, accu, orig) → TAIL(l)
REV(x, l, accu, orig) → LENGTH(orig)
REVERSE(l) → REV(0, l, nil, 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.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 5 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

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

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)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

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.
By using the subterm criterion [20] together with the size-change analysis [32] 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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

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)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

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

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)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

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

LENGTH(cons(x, l)) → LENGTH(l)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

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.
The DP Problem is simplified using the Induction Calculus [18] 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(x27, x28)=true∧length(x15)=s(x28)∧(∀x29,x30,x31:lt(x27, x28)=true∧length(x29)=x28 ⇒ IF(true, x27, x30, x31, x29)≥REV(s(x27), tail(x30), cons(head(x30), x31), x29)) ⇒ IF(true, s(x27), x13, x14, x15)≥REV(s(s(x27)), 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(x28) which results in the following new constraint:
(8)    (s(length(x38))=s(x28)∧lt(x27, x28)=true∧(∀x29,x30,x31:lt(x27, x28)=true∧length(x29)=x28 ⇒ IF(true, x27, x30, x31, x29)≥REV(s(x27), tail(x30), cons(head(x30), x31), x29))∧(∀x39,x40,x41,x42,x43,x44,x45:length(x38)=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, x38)≥REV(s(s(x40)), tail(x44), cons(head(x44), x45), x38)) ⇒ IF(true, s(x27), x13, x14, cons(x37, x38))≥REV(s(s(x27)), tail(x13), cons(head(x13), x14), cons(x37, x38)))

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

We simplified constraint (9) using rules (I), (II), (IV) which results in the following new constraint:
(10)    (IF(true, 0, x13, x14, cons(x32, x33))≥REV(s(0), tail(x13), cons(head(x13), x14), cons(x32, x33)))

We simplified constraint (8) using rules (I), (II) which results in the following new constraint:
(11)    (length(x38)=x28∧lt(x27, x28)=true∧(∀x29,x30,x31:lt(x27, x28)=true∧length(x29)=x28 ⇒ IF(true, x27, x30, x31, x29)≥REV(s(x27), tail(x30), cons(head(x30), x31), x29))∧(∀x39,x40,x41,x42,x43,x44,x45:length(x38)=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, x38)≥REV(s(s(x40)), tail(x44), cons(head(x44), x45), x38)) ⇒ IF(true, s(x27), x13, x14, cons(x37, x38))≥REV(s(s(x27)), tail(x13), cons(head(x13), x14), cons(x37, x38)))

We simplified constraint (11) using rule (VI) where we applied the induction hypothesis (∀x29,x30,x31:lt(x27, x28)=true∧length(x29)=x28 ⇒ IF(true, x27, x30, x31, x29)≥REV(s(x27), tail(x30), cons(head(x30), x31), x29)) with σ = [x31 / x14, x29 / x38, x30 / x13] which results in the following new constraint:
(12)    (IF(true, x27, x13, x14, x38)≥REV(s(x27), tail(x13), cons(head(x13), x14), x38)∧(∀x39,x40,x41,x42,x43,x44,x45:length(x38)=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, x38)≥REV(s(s(x40)), tail(x44), cons(head(x44), x45), x38)) ⇒ IF(true, s(x27), x13, x14, cons(x37, x38))≥REV(s(s(x27)), tail(x13), cons(head(x13), x14), cons(x37, x38)))

We simplified constraint (12) using rule (IV) which results in the following new constraint:
(13)    (IF(true, x27, x13, x14, x38)≥REV(s(x27), tail(x13), cons(head(x13), x14), x38) ⇒ IF(true, s(x27), x13, x14, cons(x37, x38))≥REV(s(s(x27)), tail(x13), cons(head(x13), x14), cons(x37, x38)))

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(x32, x33))≥REV(s(0), tail(x13), cons(head(x13), x14), cons(x32, x33)))
- (IF(true, x27, x13, x14, x38)≥REV(s(x27), tail(x13), cons(head(x13), x14), x38) ⇒ IF(true, s(x27), x13, x14, cons(x37, x38))≥REV(s(s(x27)), tail(x13), cons(head(x13), x14), cons(x37, x38)))

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 [18]:

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

The following pairs are in P_>:

REV(x, l, accu, orig) → IF(lt(x, length(orig)), x, 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:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ AND

                            ↳ QDP

                              ↳ DependencyGraphProof

                            ↳ QDP

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)

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.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ AND

                            ↳ QDP

                            ↳ QDP

                              ↳ DependencyGraphProof

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.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.