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)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

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)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(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:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(s(s(x0)))


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

LT(s(x), s(y)) → LT(x, y)
HALF(s(s(x))) → HALF(x)
HELP(x, y) → LT(y, x)
LOGARITHM(x) → LT(0, x)
IFA(true, x) → HELP(x, 1)
IFB(true, x, y) → HALF(x)
IFB(true, x, y) → HELP(half(x), s(y))
HELP(x, y) → IFB(lt(y, x), x, y)
LOGARITHM(x) → IFA(lt(0, x), 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)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(s(s(x0)))

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:

LT(s(x), s(y)) → LT(x, y)
HALF(s(s(x))) → HALF(x)
HELP(x, y) → LT(y, x)
LOGARITHM(x) → LT(0, x)
IFA(true, x) → HELP(x, 1)
IFB(true, x, y) → HALF(x)
IFB(true, x, y) → HELP(half(x), s(y))
HELP(x, y) → IFB(lt(y, x), x, y)
LOGARITHM(x) → IFA(lt(0, x), 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)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 5 less nodes.

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

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

HALF(s(s(x))) → HALF(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)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(s(s(x0)))

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:

HALF(s(s(x))) → HALF(x)

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

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(s(s(x0)))

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))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(s(s(x0)))



↳ 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:

HALF(s(s(x))) → HALF(x)

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:

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)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(s(s(x0)))

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:

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))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(s(s(x0)))

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))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(s(s(x0)))



↳ 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:

LT(s(x), s(y)) → LT(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

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



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

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

IFB(true, x, y) → HELP(half(x), s(y))
HELP(x, y) → IFB(lt(y, x), 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)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(s(s(x0)))

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:

IFB(true, x, y) → HELP(half(x), s(y))
HELP(x, y) → IFB(lt(y, x), x, y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
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))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(s(s(x0)))

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.

logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)



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

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

HELP(x, y) → IFB(lt(y, x), x, y)
IFB(true, x, y) → HELP(half(x), s(y))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
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))
half(0)
half(s(0))
half(s(s(x0)))

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.


IFB(true, x, y) → HELP(half(x), s(y))
The remaining pairs can at least be oriented weakly.

HELP(x, y) → IFB(lt(y, x), x, y)
Used ordering: Polynomial interpretation [25,35]:

POL(half(x1)) = (1/2)x_1   
POL(true) = 1/4   
POL(false) = 0   
POL(s(x1)) = 4 + (2)x_1   
POL(IFB(x1, x2, x3)) = (1/4)x_1 + (1/2)x_2   
POL(lt(x1, x2)) = (1/4)x_2   
POL(HELP(x1, x2)) = x_1   
POL(0) = 0   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

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



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

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

HELP(x, y) → IFB(lt(y, x), x, y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
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))
half(0)
half(s(0))
half(s(s(x0)))

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.