Termination w.r.t. Q proof of /tmp/tmp1MrjIR/thiemann24.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)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → 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)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → 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)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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

TIMES(s(x), y) → PLUS(y, times(x, y))
LT(s(x), s(y)) → LT(x, y)
FAC(x) → LOOP(x, s(0), s(0))
IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))
LOOP(x, c, y) → IF(lt(x, c), x, c, y)
IF(false, x, c, y) → TIMES(y, s(c))
LOOP(x, c, y) → LT(x, c)
TIMES(s(x), y) → TIMES(x, y)
PLUS(s(x), y) → PLUS(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)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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:

TIMES(s(x), y) → PLUS(y, times(x, y))
LT(s(x), s(y)) → LT(x, y)
FAC(x) → LOOP(x, s(0), s(0))
IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))
LOOP(x, c, y) → IF(lt(x, c), x, c, y)
IF(false, x, c, y) → TIMES(y, s(c))
LOOP(x, c, y) → LT(x, c)
TIMES(s(x), y) → TIMES(x, y)
PLUS(s(x), y) → PLUS(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)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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

PLUS(s(x), y) → PLUS(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)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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:

PLUS(s(x), y) → PLUS(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))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

↳ 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:

PLUS(s(x), y) → PLUS(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:

PLUS(s(x), y) → PLUS(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:

TIMES(s(x), y) → TIMES(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)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

↳ 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:

TIMES(s(x), y) → TIMES(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))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

↳ 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:

TIMES(s(x), y) → TIMES(x, y)

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

TIMES(s(x), y) → TIMES(x, y)
The graph contains the following edges 1 > 1, 2 >= 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:

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)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

↳ 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))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

↳ 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(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))
LOOP(x, c, y) → IF(lt(x, c), x, c, 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)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

↳ 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(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))
LOOP(x, c, y) → IF(lt(x, c), x, c, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

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.

fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

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

LOOP(x, c, y) → IF(lt(x, c), x, c, y)
IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(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 IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c))) the following chains were created:

We consider the chain LOOP(x, c, y) → IF(lt(x, c), x, c, y), IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c))) which results in the following constraint:
(1)    (IF(lt(x3, x4), x3, x4, x5)=IF(false, x6, x7, x8) ⇒ IF(false, x6, x7, x8)≥LOOP(x6, s(x7), times(x8, s(x7))))

We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
(2)    (lt(x3, x4)=false ⇒ IF(false, x3, x4, x5)≥LOOP(x3, s(x4), times(x5, s(x4))))

We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on lt(x3, x4)=false which results in the following new constraints:
(3)    (false=false ⇒ IF(false, x19, 0, x5)≥LOOP(x19, s(0), times(x5, s(0))))

(4)    (lt(x20, x21)=false∧(∀x22:lt(x20, x21)=false ⇒ IF(false, x20, x21, x22)≥LOOP(x20, s(x21), times(x22, s(x21)))) ⇒ IF(false, s(x20), s(x21), x5)≥LOOP(s(x20), s(s(x21)), times(x5, s(s(x21)))))

We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
(5)    (IF(false, x19, 0, x5)≥LOOP(x19, s(0), times(x5, s(0))))

We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (∀x22:lt(x20, x21)=false ⇒ IF(false, x20, x21, x22)≥LOOP(x20, s(x21), times(x22, s(x21)))) with σ = [x22 / x5] which results in the following new constraint:
(6)    (IF(false, x20, x21, x5)≥LOOP(x20, s(x21), times(x5, s(x21))) ⇒ IF(false, s(x20), s(x21), x5)≥LOOP(s(x20), s(s(x21)), times(x5, s(s(x21)))))

For Pair LOOP(x, c, y) → IF(lt(x, c), x, c, y) the following chains were created:

We consider the chain IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c))), LOOP(x, c, y) → IF(lt(x, c), x, c, y) which results in the following constraint:
(7) (LOOP(x9, s(x10), times(x11, s(x10)))=LOOP(x12, x13, x14) ⇒ LOOP(x12, x13, x14)≥IF(lt(x12, x13), x12, x13, x14))

We simplified constraint (7) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint:
(8) (LOOP(x9, s(x10), x14)≥IF(lt(x9, s(x10)), x9, s(x10), x14))

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

IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))
- (IF(false, x19, 0, x5)≥LOOP(x19, s(0), times(x5, s(0))))
- (IF(false, x20, x21, x5)≥LOOP(x20, s(x21), times(x5, s(x21))) ⇒ IF(false, s(x20), s(x21), x5)≥LOOP(s(x20), s(s(x21)), times(x5, s(s(x21)))))
LOOP(x, c, y) → IF(lt(x, c), x, c, y)
- (LOOP(x9, s(x10), x14)≥IF(lt(x9, s(x10)), x9, s(x10), x14))

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)) = -1 - x₁ + x₂ - x₃
POL(LOOP(x₁, x₂, x₃)) = -1 + x₁ - x₂
POL(c) = -2
POL(false) = 1
POL(lt(x₁, x₂)) = 1
POL(plus(x₁, x₂)) = 0
POL(s(x₁)) = 1 + x₁
POL(times(x₁, x₂)) = 0
POL(true) = 1

The following pairs are in P_>:

LOOP(x, c, y) → IF(lt(x, c), x, c, y)

The following pairs are in P_bound:

IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))

The following rules are usable:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ AND

                            ↳ QDP

                              ↳ DependencyGraphProof

                            ↳ QDP

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

IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ AND

                            ↳ QDP

                            ↳ QDP

                              ↳ DependencyGraphProof

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

LOOP(x, c, y) → IF(lt(x, c), x, c, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(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.