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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_mod(true, x, y) → mod(minus(x, y), y)
if_mod(false, s(x), s(y)) → s(x)

Q is empty.


QTRS
  ↳ Overlay + Local Confluence

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_mod(true, x, y) → mod(minus(x, y), y)
if_mod(false, s(x), s(y)) → s(x)

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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_mod(true, x, y) → mod(minus(x, y), y)
if_mod(false, s(x), s(y)) → s(x)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, s(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:

IF_MOD(true, x, y) → MOD(minus(x, y), y)
MINUS(s(x), s(y)) → MINUS(x, y)
IF_MOD(true, x, y) → MINUS(x, y)
MOD(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y))
MOD(s(x), s(y)) → LE(y, x)
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_mod(true, x, y) → mod(minus(x, y), y)
if_mod(false, s(x), s(y)) → s(x)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, s(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:

IF_MOD(true, x, y) → MOD(minus(x, y), y)
MINUS(s(x), s(y)) → MINUS(x, y)
IF_MOD(true, x, y) → MINUS(x, y)
MOD(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y))
MOD(s(x), s(y)) → LE(y, x)
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_mod(true, x, y) → mod(minus(x, y), y)
if_mod(false, s(x), s(y)) → s(x)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 2 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:

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_mod(true, x, y) → mod(minus(x, y), y)
if_mod(false, s(x), s(y)) → s(x)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, s(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

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

MINUS(s(x), s(y)) → MINUS(x, y)

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, s(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.

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, s(x0), s(x1))



↳ 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:

MINUS(s(x), s(y)) → MINUS(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:



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP
                ↳ UsableRulesProof
              ↳ QDP

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

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_mod(true, x, y) → mod(minus(x, y), y)
if_mod(false, s(x), s(y)) → s(x)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, s(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
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof
              ↳ QDP

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

LE(s(x), s(y)) → LE(x, y)

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, s(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.

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, s(x0), s(x1))



↳ 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:

LE(s(x), s(y)) → LE(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:



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
QDP
                ↳ UsableRulesProof

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

IF_MOD(true, x, y) → MOD(minus(x, y), y)
MOD(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))
if_mod(true, x, y) → mod(minus(x, y), y)
if_mod(false, s(x), s(y)) → s(x)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, s(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
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof

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

IF_MOD(true, x, y) → MOD(minus(x, y), y)
MOD(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, s(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.

mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))
if_mod(true, x0, x1)
if_mod(false, s(x0), s(x1))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
QDP
                        ↳ Narrowing

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

IF_MOD(true, x, y) → MOD(minus(x, y), y)
MOD(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule IF_MOD(true, x, y) → MOD(minus(x, y), y) at position [0] we obtained the following new rules:

IF_MOD(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(x1))
IF_MOD(true, x0, 0) → MOD(x0, 0)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
QDP
                            ↳ DependencyGraphProof

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

IF_MOD(true, x0, 0) → MOD(x0, 0)
MOD(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y))
IF_MOD(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(x1))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
QDP
                                ↳ Narrowing

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

MOD(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y))
IF_MOD(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(x1))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule MOD(s(x), s(y)) → IF_MOD(le(y, x), s(x), s(y)) at position [0] we obtained the following new rules:

MOD(s(x0), s(0)) → IF_MOD(true, s(x0), s(0))
MOD(s(0), s(s(x0))) → IF_MOD(false, s(0), s(s(x0)))
MOD(s(s(x1)), s(s(x0))) → IF_MOD(le(x0, x1), s(s(x1)), s(s(x0)))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
QDP
                                    ↳ DependencyGraphProof

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

MOD(s(x0), s(0)) → IF_MOD(true, s(x0), s(0))
MOD(s(s(x1)), s(s(x0))) → IF_MOD(le(x0, x1), s(s(x1)), s(s(x0)))
MOD(s(0), s(s(x0))) → IF_MOD(false, s(0), s(s(x0)))
IF_MOD(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(x1))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ DependencyGraphProof
QDP
                                        ↳ Instantiation

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

MOD(s(x0), s(0)) → IF_MOD(true, s(x0), s(0))
MOD(s(s(x1)), s(s(x0))) → IF_MOD(le(x0, x1), s(s(x1)), s(s(x0)))
IF_MOD(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(x1))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF_MOD(true, s(x0), s(x1)) → MOD(minus(x0, x1), s(x1)) we obtained the following new rules:

IF_MOD(true, s(s(z0)), s(s(z1))) → MOD(minus(s(z0), s(z1)), s(s(z1)))
IF_MOD(true, s(z0), s(0)) → MOD(minus(z0, 0), s(0))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ Instantiation
QDP
                                            ↳ DependencyGraphProof

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

MOD(s(x0), s(0)) → IF_MOD(true, s(x0), s(0))
MOD(s(s(x1)), s(s(x0))) → IF_MOD(le(x0, x1), s(s(x1)), s(s(x0)))
IF_MOD(true, s(s(z0)), s(s(z1))) → MOD(minus(s(z0), s(z1)), s(s(z1)))
IF_MOD(true, s(z0), s(0)) → MOD(minus(z0, 0), s(0))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ Instantiation
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ AND
QDP
                                                  ↳ Rewriting
                                                ↳ QDP

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

MOD(s(s(x1)), s(s(x0))) → IF_MOD(le(x0, x1), s(s(x1)), s(s(x0)))
IF_MOD(true, s(s(z0)), s(s(z1))) → MOD(minus(s(z0), s(z1)), s(s(z1)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))

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

IF_MOD(true, s(s(z0)), s(s(z1))) → MOD(minus(z0, z1), s(s(z1)))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ Instantiation
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ AND
                                                ↳ QDP
                                                  ↳ Rewriting
QDP
                                                      ↳ QDPOrderProof
                                                      ↳ QDPOrderProof
                                                ↳ QDP

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

MOD(s(s(x1)), s(s(x0))) → IF_MOD(le(x0, x1), s(s(x1)), s(s(x0)))
IF_MOD(true, s(s(z0)), s(s(z1))) → MOD(minus(z0, z1), s(s(z1)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


IF_MOD(true, s(s(z0)), s(s(z1))) → MOD(minus(z0, z1), s(s(z1)))
The remaining pairs can at least be oriented weakly.

MOD(s(s(x1)), s(s(x0))) → IF_MOD(le(x0, x1), s(s(x1)), s(s(x0)))
Used ordering: Polynomial interpretation with max and min functions [25]:

POL(0) = 0   
POL(IF_MOD(x1, x2, x3)) = x2 + x3   
POL(MOD(x1, x2)) = x1 + x2   
POL(false) = 0   
POL(le(x1, x2)) = 0   
POL(minus(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

The following usable rules [17] were oriented:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ Instantiation
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ AND
                                                ↳ QDP
                                                  ↳ Rewriting
                                                    ↳ QDP
                                                      ↳ QDPOrderProof
QDP
                                                          ↳ DependencyGraphProof
                                                      ↳ QDPOrderProof
                                                ↳ QDP

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

MOD(s(s(x1)), s(s(x0))) → IF_MOD(le(x0, x1), s(s(x1)), s(s(x0)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(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.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


MOD(s(s(x1)), s(s(x0))) → IF_MOD(le(x0, x1), s(s(x1)), s(s(x0)))
IF_MOD(true, s(s(z0)), s(s(z1))) → MOD(minus(z0, z1), s(s(z1)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:

POL(0) = 0   
POL(IF_MOD(x1, x2, x3)) = x2 + x3   
POL(MOD(x1, x2)) = 1 + x1 + x2   
POL(false) = 0   
POL(le(x1, x2)) = 0   
POL(minus(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

The following usable rules [17] were oriented:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ Instantiation
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ AND
                                                ↳ QDP
                                                  ↳ Rewriting
                                                    ↳ QDP
                                                      ↳ QDPOrderProof
                                                      ↳ QDPOrderProof
QDP
                                                ↳ QDP

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ Instantiation
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ AND
                                                ↳ QDP
QDP
                                                  ↳ UsableRulesProof

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

MOD(s(x0), s(0)) → IF_MOD(true, s(x0), s(0))
IF_MOD(true, s(z0), s(0)) → MOD(minus(z0, 0), s(0))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(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
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ Instantiation
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ AND
                                                ↳ QDP
                                                ↳ QDP
                                                  ↳ UsableRulesProof
QDP
                                                      ↳ QReductionProof

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

MOD(s(x0), s(0)) → IF_MOD(true, s(x0), s(0))
IF_MOD(true, s(z0), s(0)) → MOD(minus(z0, 0), s(0))

The TRS R consists of the following rules:

minus(x, 0) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(s(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.

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ Instantiation
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ AND
                                                ↳ QDP
                                                ↳ QDP
                                                  ↳ UsableRulesProof
                                                    ↳ QDP
                                                      ↳ QReductionProof
QDP
                                                          ↳ Rewriting

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

MOD(s(x0), s(0)) → IF_MOD(true, s(x0), s(0))
IF_MOD(true, s(z0), s(0)) → MOD(minus(z0, 0), s(0))

The TRS R consists of the following rules:

minus(x, 0) → x

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))

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

IF_MOD(true, s(z0), s(0)) → MOD(z0, s(0))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ Instantiation
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ AND
                                                ↳ QDP
                                                ↳ QDP
                                                  ↳ UsableRulesProof
                                                    ↳ QDP
                                                      ↳ QReductionProof
                                                        ↳ QDP
                                                          ↳ Rewriting
QDP
                                                              ↳ UsableRulesProof

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

MOD(s(x0), s(0)) → IF_MOD(true, s(x0), s(0))
IF_MOD(true, s(z0), s(0)) → MOD(z0, s(0))

The TRS R consists of the following rules:

minus(x, 0) → x

The set Q consists of the following terms:

minus(x0, 0)
minus(s(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
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ Instantiation
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ AND
                                                ↳ QDP
                                                ↳ QDP
                                                  ↳ UsableRulesProof
                                                    ↳ QDP
                                                      ↳ QReductionProof
                                                        ↳ QDP
                                                          ↳ Rewriting
                                                            ↳ QDP
                                                              ↳ UsableRulesProof
QDP
                                                                  ↳ QReductionProof

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

MOD(s(x0), s(0)) → IF_MOD(true, s(x0), s(0))
IF_MOD(true, s(z0), s(0)) → MOD(z0, s(0))

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

minus(x0, 0)
minus(s(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.

minus(x0, 0)
minus(s(x0), s(x1))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ Instantiation
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ AND
                                                ↳ QDP
                                                ↳ QDP
                                                  ↳ UsableRulesProof
                                                    ↳ QDP
                                                      ↳ QReductionProof
                                                        ↳ QDP
                                                          ↳ Rewriting
                                                            ↳ QDP
                                                              ↳ UsableRulesProof
                                                                ↳ QDP
                                                                  ↳ QReductionProof
QDP
                                                                      ↳ ForwardInstantiation

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

MOD(s(x0), s(0)) → IF_MOD(true, s(x0), s(0))
IF_MOD(true, s(z0), s(0)) → MOD(z0, s(0))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule IF_MOD(true, s(z0), s(0)) → MOD(z0, s(0)) we obtained the following new rules:

IF_MOD(true, s(s(y_0)), s(0)) → MOD(s(y_0), s(0))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ Instantiation
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ AND
                                                ↳ QDP
                                                ↳ QDP
                                                  ↳ UsableRulesProof
                                                    ↳ QDP
                                                      ↳ QReductionProof
                                                        ↳ QDP
                                                          ↳ Rewriting
                                                            ↳ QDP
                                                              ↳ UsableRulesProof
                                                                ↳ QDP
                                                                  ↳ QReductionProof
                                                                    ↳ QDP
                                                                      ↳ ForwardInstantiation
QDP
                                                                          ↳ ForwardInstantiation

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

MOD(s(x0), s(0)) → IF_MOD(true, s(x0), s(0))
IF_MOD(true, s(s(y_0)), s(0)) → MOD(s(y_0), s(0))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule MOD(s(x0), s(0)) → IF_MOD(true, s(x0), s(0)) we obtained the following new rules:

MOD(s(s(y_0)), s(0)) → IF_MOD(true, s(s(y_0)), s(0))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ Instantiation
                                          ↳ QDP
                                            ↳ DependencyGraphProof
                                              ↳ AND
                                                ↳ QDP
                                                ↳ QDP
                                                  ↳ UsableRulesProof
                                                    ↳ QDP
                                                      ↳ QReductionProof
                                                        ↳ QDP
                                                          ↳ Rewriting
                                                            ↳ QDP
                                                              ↳ UsableRulesProof
                                                                ↳ QDP
                                                                  ↳ QReductionProof
                                                                    ↳ QDP
                                                                      ↳ ForwardInstantiation
                                                                        ↳ QDP
                                                                          ↳ ForwardInstantiation
QDP
                                                                              ↳ QDPSizeChangeProof

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

MOD(s(s(y_0)), s(0)) → IF_MOD(true, s(s(y_0)), s(0))
IF_MOD(true, s(s(y_0)), s(0)) → MOD(s(y_0), s(0))

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: