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
  ↳ 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))

Q is empty.

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))

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

↳ 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))

Q is empty.
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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ 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))

Q is empty.
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.


HALF(s(s(x))) → HALF(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(HALF(x1)) = x_1   
POL(s(x1)) = 5/4 + (5/2)x_1   
The value of delta used in the strict ordering is 35/8.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ 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))

Q is empty.
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.


LT(s(x), s(y)) → LT(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x1)) = 1/2 + (13/4)x_1   
POL(LT(x1, x2)) = (15/4)x_2   
The value of delta used in the strict ordering is 15/8.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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))

Q is empty.
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)) = (1/4)x_1   
POL(true) = 1/4   
POL(false) = 0   
POL(s(x1)) = 1/2 + (4)x_1   
POL(IFB(x1, x2, x3)) = (1/2)x_1 + x_2   
POL(lt(x1, x2)) = (2)x_2   
POL(HELP(x1, x2)) = (4)x_1   
POL(0) = 0   
The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ 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:

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.
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.