Termination w.r.t. Q proof of /tmp/tmpiz8rlV/otto04.trs

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:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))

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:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(x1))

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

FIBO(s(s(x))) → FIBO(x)
FIBO(s(0)) → FIB(s(0))
IF(true, c, s(s(x)), a, b) → IF(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
FIBO(0) → FIB(0)
FIBO(s(s(x))) → FIBO(s(x))
IF(false, c, s(s(x)), a, b) → SUM(fibo(a), fibo(b))
LT(s(x), s(y)) → LT(x, y)
FIB(s(s(x))) → IF(true, 0, s(s(x)), 0, 0)
FIBO(s(s(x))) → SUM(fibo(s(x)), fibo(x))
IF(false, c, s(s(x)), a, b) → FIBO(b)
SUM(x, s(y)) → SUM(x, y)
IF(false, c, s(s(x)), a, b) → FIBO(a)
IF(true, c, s(s(x)), a, b) → LT(s(c), s(s(x)))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(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:

FIBO(s(s(x))) → FIBO(x)
FIBO(s(0)) → FIB(s(0))
IF(true, c, s(s(x)), a, b) → IF(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
FIBO(0) → FIB(0)
FIBO(s(s(x))) → FIBO(s(x))
IF(false, c, s(s(x)), a, b) → SUM(fibo(a), fibo(b))
LT(s(x), s(y)) → LT(x, y)
FIB(s(s(x))) → IF(true, 0, s(s(x)), 0, 0)
FIBO(s(s(x))) → SUM(fibo(s(x)), fibo(x))
IF(false, c, s(s(x)), a, b) → FIBO(b)
SUM(x, s(y)) → SUM(x, y)
IF(false, c, s(s(x)), a, b) → FIBO(a)
IF(true, c, s(s(x)), a, b) → LT(s(c), s(s(x)))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 8 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:

SUM(x, s(y)) → SUM(x, y)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(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 [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

              ↳ QDP

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

SUM(x, s(y)) → SUM(x, y)

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

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(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.

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(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:

SUM(x, s(y)) → SUM(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:

SUM(x, s(y)) → SUM(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:

FIBO(s(s(x))) → FIBO(x)
FIBO(s(s(x))) → FIBO(s(x))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(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:

FIBO(s(s(x))) → FIBO(x)
FIBO(s(s(x))) → FIBO(s(x))

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

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(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.

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(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:

FIBO(s(s(x))) → FIBO(x)
FIBO(s(s(x))) → FIBO(s(x))

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

FIBO(s(s(x))) → FIBO(x)
The graph contains the following edges 1 > 1
FIBO(s(s(x))) → FIBO(s(x))
The graph contains the following edges 1 > 1

↳ 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:

LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(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:

LT(s(x), s(y)) → LT(x, y)

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

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(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.

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(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:

LT(s(x), s(y)) → LT(x, y)

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

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

IF(true, c, s(s(x)), a, b) → IF(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fibo(0) → fib(0)
fibo(s(0)) → fib(s(0))
fibo(s(s(x))) → sum(fibo(s(x)), fibo(x))
fib(0) → s(0)
fib(s(0)) → s(0)
fib(s(s(x))) → if(true, 0, s(s(x)), 0, 0)
if(true, c, s(s(x)), a, b) → if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
if(false, c, s(s(x)), a, b) → sum(fibo(a), fibo(b))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(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:

IF(true, c, s(s(x)), a, b) → IF(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(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.

fibo(0)
fibo(s(0))
fibo(s(s(x0)))
fib(0)
fib(s(0))
fib(s(s(x0)))
if(true, x0, s(s(x1)), x2, x3)
if(false, x0, s(s(x1)), x2, x3)
sum(x0, 0)
sum(x0, s(x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

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

IF(true, c, s(s(x)), a, b) → IF(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule IF(true, c, s(s(x)), a, b) → IF(lt(s(c), s(s(x))), s(c), s(s(x)), b, c) at position [0] we obtained the following new rules:

IF(true, c, s(s(x)), a, b) → IF(lt(c, s(x)), s(c), s(s(x)), b, c)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Instantiation

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

IF(true, c, s(s(x)), a, b) → IF(lt(c, s(x)), s(c), s(s(x)), b, c)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF(true, c, s(s(x)), a, b) → IF(lt(c, s(x)), s(c), s(s(x)), b, c) we obtained the following new rules:

IF(true, s(z0), s(s(z1)), z3, z0) → IF(lt(s(z0), s(z1)), s(s(z0)), s(s(z1)), z0, s(z0))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ Rewriting

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

IF(true, s(z0), s(s(z1)), z3, z0) → IF(lt(s(z0), s(z1)), s(s(z0)), s(s(z1)), z0, s(z0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule IF(true, s(z0), s(s(z1)), z3, z0) → IF(lt(s(z0), s(z1)), s(s(z0)), s(s(z1)), z0, s(z0)) at position [0] we obtained the following new rules:

IF(true, s(z0), s(s(z1)), z3, z0) → IF(lt(z0, z1), s(s(z0)), s(s(z1)), z0, s(z0))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Instantiation

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

IF(true, s(z0), s(s(z1)), z3, z0) → IF(lt(z0, z1), s(s(z0)), s(s(z1)), z0, s(z0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF(true, s(z0), s(s(z1)), z3, z0) → IF(lt(z0, z1), s(s(z0)), s(s(z1)), z0, s(z0)) we obtained the following new rules:

IF(true, s(s(z0)), s(s(z1)), z0, s(z0)) → IF(lt(s(z0), z1), s(s(s(z0))), s(s(z1)), s(z0), s(s(z0)))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ NonInfProof

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

IF(true, s(s(z0)), s(s(z1)), z0, s(z0)) → IF(lt(s(z0), z1), s(s(s(z0))), s(s(z1)), s(z0), s(s(z0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(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 IF(true, s(s(z0)), s(s(z1)), z0, s(z0)) → IF(lt(s(z0), z1), s(s(s(z0))), s(s(z1)), s(z0), s(s(z0))) the following chains were created:

We consider the chain IF(true, s(s(z0)), s(s(z1)), z0, s(z0)) → IF(lt(s(z0), z1), s(s(s(z0))), s(s(z1)), s(z0), s(s(z0))), IF(true, s(s(z0)), s(s(z1)), z0, s(z0)) → IF(lt(s(z0), z1), s(s(s(z0))), s(s(z1)), s(z0), s(s(z0))) which results in the following constraint:
(1)    (IF(lt(s(x0), x1), s(s(s(x0))), s(s(x1)), s(x0), s(s(x0)))=IF(true, s(s(x2)), s(s(x3)), x2, s(x2)) ⇒ IF(true, s(s(x2)), s(s(x3)), x2, s(x2))≥IF(lt(s(x2), x3), s(s(s(x2))), s(s(x3)), s(x2), s(s(x2))))

We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint:
(2)    (s(x0)=x4∧lt(x4, x1)=true ⇒ IF(true, s(s(s(x0))), s(s(x1)), s(x0), s(s(x0)))≥IF(lt(s(s(x0)), x1), s(s(s(s(x0)))), s(s(x1)), s(s(x0)), s(s(s(x0)))))

We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on lt(x4, x1)=true which results in the following new constraints:
(3)    (lt(x5, x6)=true∧s(x0)=s(x5)∧(∀x7:lt(x5, x6)=true∧s(x7)=x5 ⇒ IF(true, s(s(s(x7))), s(s(x6)), s(x7), s(s(x7)))≥IF(lt(s(s(x7)), x6), s(s(s(s(x7)))), s(s(x6)), s(s(x7)), s(s(s(x7))))) ⇒ IF(true, s(s(s(x0))), s(s(s(x6))), s(x0), s(s(x0)))≥IF(lt(s(s(x0)), s(x6)), s(s(s(s(x0)))), s(s(s(x6))), s(s(x0)), s(s(s(x0)))))

(4)    (true=true∧s(x0)=0 ⇒ IF(true, s(s(s(x0))), s(s(s(x8))), s(x0), s(s(x0)))≥IF(lt(s(s(x0)), s(x8)), s(s(s(s(x0)))), s(s(s(x8))), s(s(x0)), s(s(s(x0)))))

We simplified constraint (3) using rules (I), (II), (III), (IV) which results in the following new constraint:
(5)    (lt(x5, x6)=true ⇒ IF(true, s(s(s(x5))), s(s(s(x6))), s(x5), s(s(x5)))≥IF(lt(s(s(x5)), s(x6)), s(s(s(s(x5)))), s(s(s(x6))), s(s(x5)), s(s(s(x5)))))

We solved constraint (4) using rules (I), (II).We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on lt(x5, x6)=true which results in the following new constraints:
(6)    (lt(x10, x11)=true∧(lt(x10, x11)=true ⇒ IF(true, s(s(s(x10))), s(s(s(x11))), s(x10), s(s(x10)))≥IF(lt(s(s(x10)), s(x11)), s(s(s(s(x10)))), s(s(s(x11))), s(s(x10)), s(s(s(x10))))) ⇒ IF(true, s(s(s(s(x10)))), s(s(s(s(x11)))), s(s(x10)), s(s(s(x10))))≥IF(lt(s(s(s(x10))), s(s(x11))), s(s(s(s(s(x10))))), s(s(s(s(x11)))), s(s(s(x10))), s(s(s(s(x10))))))

(7)    (true=true ⇒ IF(true, s(s(s(0))), s(s(s(s(x12)))), s(0), s(s(0)))≥IF(lt(s(s(0)), s(s(x12))), s(s(s(s(0)))), s(s(s(s(x12)))), s(s(0)), s(s(s(0)))))

We simplified constraint (6) using rule (VI) where we applied the induction hypothesis (lt(x10, x11)=true ⇒ IF(true, s(s(s(x10))), s(s(s(x11))), s(x10), s(s(x10)))≥IF(lt(s(s(x10)), s(x11)), s(s(s(s(x10)))), s(s(s(x11))), s(s(x10)), s(s(s(x10))))) with σ = [ ] which results in the following new constraint:
(8)    (IF(true, s(s(s(x10))), s(s(s(x11))), s(x10), s(s(x10)))≥IF(lt(s(s(x10)), s(x11)), s(s(s(s(x10)))), s(s(s(x11))), s(s(x10)), s(s(s(x10)))) ⇒ IF(true, s(s(s(s(x10)))), s(s(s(s(x11)))), s(s(x10)), s(s(s(x10))))≥IF(lt(s(s(s(x10))), s(s(x11))), s(s(s(s(s(x10))))), s(s(s(s(x11)))), s(s(s(x10))), s(s(s(s(x10))))))

We simplified constraint (7) using rules (I), (II) which results in the following new constraint:
(9)    (IF(true, s(s(s(0))), s(s(s(s(x12)))), s(0), s(s(0)))≥IF(lt(s(s(0)), s(s(x12))), s(s(s(s(0)))), s(s(s(s(x12)))), s(s(0)), s(s(s(0)))))

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

IF(true, s(s(z0)), s(s(z1)), z0, s(z0)) → IF(lt(s(z0), z1), s(s(s(z0))), s(s(z1)), s(z0), s(s(z0)))
- (IF(true, s(s(s(x10))), s(s(s(x11))), s(x10), s(s(x10)))≥IF(lt(s(s(x10)), s(x11)), s(s(s(s(x10)))), s(s(s(x11))), s(s(x10)), s(s(s(x10)))) ⇒ IF(true, s(s(s(s(x10)))), s(s(s(s(x11)))), s(s(x10)), s(s(s(x10))))≥IF(lt(s(s(s(x10))), s(s(x11))), s(s(s(s(s(x10))))), s(s(s(s(x11)))), s(s(s(x10))), s(s(s(s(x10))))))
- (IF(true, s(s(s(0))), s(s(s(s(x12)))), s(0), s(s(0)))≥IF(lt(s(s(0)), s(s(x12))), s(s(s(s(0)))), s(s(s(s(x12)))), s(s(0)), s(s(s(0)))))

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)) = -1 - x₁ - x₂ + x₃ + x₄ - x₅
POL(c) = -1
POL(false) = 1
POL(lt(x₁, x₂)) = 0
POL(s(x₁)) = 1 + x₁
POL(true) = 0

The following pairs are in P_>:

IF(true, s(s(z0)), s(s(z1)), z0, s(z0)) → IF(lt(s(z0), z1), s(s(s(z0))), s(s(z1)), s(z0), s(s(z0)))

The following pairs are in P_bound:

IF(true, s(s(z0)), s(s(z1)), z0, s(z0)) → IF(lt(s(z0), z1), s(s(s(z0))), s(s(z1)), s(z0), s(s(z0)))

The following rules are usable:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ NonInfProof

                                          ↳ QDP

                                            ↳ PisEmptyProof

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

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(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.