Termination w.r.t. Q proof of /tmp/tmp4rxE0c/parting01

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:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

Q restricted rewrite system:
The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

Q is empty.

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

LAST(cons(x, cons(y, xs))) → LAST(cons(y, xs))
DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)
DEL(x, cons(y, xs)) → EQ(x, y)
IF(false, x, y, xs) → DEL(x, xs)
EQ(s(x), s(y)) → EQ(x, y)
REVERSE(cons(x, xs)) → LAST(cons(x, xs))
REVERSE(cons(x, xs)) → REVERSE(del(last(cons(x, xs)), cons(x, xs)))
REVERSE(cons(x, xs)) → DEL(last(cons(x, xs)), cons(x, xs))

The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

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:

LAST(cons(x, cons(y, xs))) → LAST(cons(y, xs))
DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)
DEL(x, cons(y, xs)) → EQ(x, y)
IF(false, x, y, xs) → DEL(x, xs)
EQ(s(x), s(y)) → EQ(x, y)
REVERSE(cons(x, xs)) → LAST(cons(x, xs))
REVERSE(cons(x, xs)) → REVERSE(del(last(cons(x, xs)), cons(x, xs)))
REVERSE(cons(x, xs)) → DEL(last(cons(x, xs)), cons(x, xs))

The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 3 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

EQ(s(x), s(y)) → EQ(x, y)

The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

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 [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

EQ(s(x), s(y)) → EQ(x, y)

R is empty.
The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

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.[THIEMANN].

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

EQ(s(x), s(y)) → EQ(x, y)

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

EQ(s(x), s(y)) → EQ(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

              ↳ QDP

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

IF(false, x, y, xs) → DEL(x, xs)
DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)

The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

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

IF(false, x, y, xs) → DEL(x, xs)
DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

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.[THIEMANN].

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
reverse(nil)
reverse(cons(x0, x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

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

IF(false, x, y, xs) → DEL(x, xs)
DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

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

DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)
The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4
IF(false, x, y, xs) → DEL(x, xs)
The graph contains the following edges 2 >= 1, 4 >= 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

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

LAST(cons(x, cons(y, xs))) → LAST(cons(y, xs))

The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

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

LAST(cons(x, cons(y, xs))) → LAST(cons(y, xs))

R is empty.
The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

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.[THIEMANN].

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

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

LAST(cons(x, cons(y, xs))) → LAST(cons(y, xs))

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

LAST(cons(x, cons(y, xs))) → LAST(cons(y, xs))
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

REVERSE(cons(x, xs)) → REVERSE(del(last(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
reverse(nil) → nil
reverse(cons(x, xs)) → cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

REVERSE(cons(x, xs)) → REVERSE(del(last(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
del(x, nil) → nil

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
reverse(nil)
reverse(cons(x0, x1))

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.[THIEMANN].

reverse(nil)
reverse(cons(x0, x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Induction-Processor

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

REVERSE(cons(x, xs)) → REVERSE(del(last(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
del(x, nil) → nil

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

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

This DP could be deleted by the Induction-Processor:
REVERSE(cons(x', xs')) → REVERSE(del(last(cons(x', xs')), cons(x', xs')))

This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 2
POL(REVERSE(x₁)) = x₁
POL(cons(x₁, x₂)) = 1 + x₁ + 2·x₂
POL(del(x₁, x₂)) = x₂
POL(eq(x₁, x₂)) = 2·x₂
POL(false) = 1
POL(if(x₁, x₂, x₃, x₄)) = 1 + x₃ + 2·x₄
POL(last(x₁)) = 3 + 2·x₁
POL(nil) = 0
POL(s(x₁)) = 1 + x₁
POL(true) = 2

At least one of these decreasing rules is always used after the deleted DP:
if(true, x595, y445, xs285) → xs285

The following formula is valid:
z0:sort[a24].(¬(z0 =nil)→del'(last(z0 ), z0 )=true)

The transformed set:
del'(x16, cons(y11, xs7)) → if'(eq(x16, y11), x16, y11, xs7)
if'(true, x59, y44, xs28) → true
if'(false, x68, y51, xs33) → del'(x68, xs33)
del'(x77, nil) → false
last(cons(x', nil)) → x'
last(cons(x7, cons(y4, xs2))) → last(cons(y4, xs2))
del(x16, cons(y11, xs7)) → if(eq(x16, y11), x16, y11, xs7)
eq(0, 0) → true
eq(0, s(y24)) → false
eq(s(x41), 0) → false
eq(s(x50), s(y37)) → eq(x50, y37)
if(true, x59, y44, xs28) → xs28
if(false, x68, y51, xs33) → cons(y51, del(x68, xs33))
del(x77, nil) → nil
last(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a24](nil, nil) → true
equal_sort[a24](nil, cons(x0, x1)) → false
equal_sort[a24](cons(x0, x1), nil) → false
equal_sort[a24](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a24](x0, x2), equal_sort[a24](x1, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42]) → true

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Induction-Processor

                          ↳ AND

                            ↳ QDP

                              ↳ PisEmptyProof

                            ↳ QTRS

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

last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
del(x, nil) → nil

The set Q consists of the following terms:

last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Induction-Processor

                          ↳ AND

                            ↳ QDP

                            ↳ QTRS

                              ↳ QTRSRRRProof

Q restricted rewrite system:
The TRS R consists of the following rules:

del'(x16, cons(y11, xs7)) → if'(eq(x16, y11), x16, y11, xs7)
if'(true, x59, y44, xs28) → true
if'(false, x68, y51, xs33) → del'(x68, xs33)
del'(x77, nil) → false
last(cons(x', nil)) → x'
last(cons(x7, cons(y4, xs2))) → last(cons(y4, xs2))
del(x16, cons(y11, xs7)) → if(eq(x16, y11), x16, y11, xs7)
eq(0, 0) → true
eq(0, s(y24)) → false
eq(s(x41), 0) → false
eq(s(x50), s(y37)) → eq(x50, y37)
if(true, x59, y44, xs28) → xs28
if(false, x68, y51, xs33) → cons(y51, del(x68, xs33))
del(x77, nil) → nil
last(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a24](nil, nil) → true
equal_sort[a24](nil, cons(x0, x1)) → false
equal_sort[a24](cons(x0, x1), nil) → false
equal_sort[a24](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a24](x0, x2), equal_sort[a24](x1, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42]) → true

Q is empty.

The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:

del'(x16, cons(y11, xs7)) → if'(eq(x16, y11), x16, y11, xs7)
if'(true, x59, y44, xs28) → true
if'(false, x68, y51, xs33) → del'(x68, xs33)
del'(x77, nil) → false
last(cons(x', nil)) → x'
last(cons(x7, cons(y4, xs2))) → last(cons(y4, xs2))
del(x16, cons(y11, xs7)) → if(eq(x16, y11), x16, y11, xs7)
eq(0, 0) → true
eq(0, s(y24)) → false
eq(s(x41), 0) → false
eq(s(x50), s(y37)) → eq(x50, y37)
if(true, x59, y44, xs28) → xs28
if(false, x68, y51, xs33) → cons(y51, del(x68, xs33))
del(x77, nil) → nil
last(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a24](nil, nil) → true
equal_sort[a24](nil, cons(x0, x1)) → false
equal_sort[a24](cons(x0, x1), nil) → false
equal_sort[a24](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a24](x0, x2), equal_sort[a24](x1, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42]) → true

Q is empty.
Used ordering:
Recursive path order with status [RPO].
Quasi-Precedence:

s₁ > eq₂
s₁ > equal_sort[a0]₂
not₁ > [del'₂, if'₄, false, nil, 0, equal_sort[a24]₂] > true
not₁ > [del'₂, if'₄, false, nil, 0, equal_sort[a24]₂] > [del₂, if₄] > [cons₂, last₁, and₂] > eq₂
isa_false₁ > [del'₂, if'₄, false, nil, 0, equal_sort[a24]₂] > true
isa_false₁ > [del'₂, if'₄, false, nil, 0, equal_sort[a24]₂] > [del₂, if₄] > [cons₂, last₁, and₂] > eq₂
equal_sort[a42]₂ > true

Status:

witness_sort[a42]: multiset
if₄: [2,4,3,1]
eq₂: [2,1]
if'₄: [2,4,3,1]
last₁: [1]
true: multiset
or₂: [2,1]
equal_sort[a42]₂: multiset
and₂: multiset
del₂: [1,2]
0: multiset
del'₂: [1,2]
equal_bool₂: [1,2]
equal_sort[a0]₂: [1,2]
cons₂: [1,2]
equal_sort[a24]₂: multiset
not₁: multiset
isa_false₁: [1]
false: multiset
s₁: [1]
nil: multiset
isa_true₁: [1]

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

del'(x16, cons(y11, xs7)) → if'(eq(x16, y11), x16, y11, xs7)
if'(true, x59, y44, xs28) → true
if'(false, x68, y51, xs33) → del'(x68, xs33)
del'(x77, nil) → false
last(cons(x', nil)) → x'
last(cons(x7, cons(y4, xs2))) → last(cons(y4, xs2))
del(x16, cons(y11, xs7)) → if(eq(x16, y11), x16, y11, xs7)
eq(0, 0) → true
eq(0, s(y24)) → false
eq(s(x41), 0) → false
eq(s(x50), s(y37)) → eq(x50, y37)
if(true, x59, y44, xs28) → xs28
if(false, x68, y51, xs33) → cons(y51, del(x68, xs33))
del(x77, nil) → nil
last(nil) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a24](nil, nil) → true
equal_sort[a24](nil, cons(x0, x1)) → false
equal_sort[a24](cons(x0, x1), nil) → false
equal_sort[a24](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a24](x0, x2), equal_sort[a24](x1, x3))
equal_sort[a42](witness_sort[a42], witness_sort[a42]) → true

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Induction-Processor

                          ↳ AND

                            ↳ QDP

                            ↳ QTRS

                              ↳ QTRSRRRProof

                                ↳ QTRS

                                  ↳ RisEmptyProof

                                  ↳ RisEmptyProof

                                  ↳ RisEmptyProof

Q restricted rewrite system:
R is empty.
Q is empty.

The TRS R is empty. Hence, termination is trivially proven.
The TRS R is empty. Hence, termination is trivially proven.
The TRS R is empty. Hence, termination is trivially proven.