Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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)))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

LT(s(x), s(y)) → LT(x, y)
LOGARITHM(x) → IFA(lt(0, x), x)
LOGARITHM(x) → LT(0, x)
IFA(true, x) → HELP(x, 1)
HELP(x, y) → IFB(lt(y, x), x, y)
HELP(x, y) → LT(y, x)
IFB(true, x, y) → HELP(half(x), s(y))
IFB(true, x, y) → HALF(x)
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.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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.

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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.

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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)))

(11) Obligation:

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.

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

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

(13) TRUE

(14) Obligation:

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.

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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.

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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)))

(18) Obligation:

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.

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

• LT(s(x), s(y)) → LT(x, y)
  The graph contains the following edges 1 > 1, 2 > 2

(20) TRUE

(21) Obligation:

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:

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.

(22) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(23) Obligation:

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))
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.

(24) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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

(25) Obligation:

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.

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(HELP(x₁, x₂)) = (2)x₁   
POL(IFB(x₁, x₂, x₃)) = (4)x₁ + (1/2)x₂   
POL(lt(x₁, x₂)) = (1/4)x₂   
POL(true) = 1   
POL(half(x₁)) = (1/4)x₁   
POL(s(x₁)) = 4 + (3)x₁   
POL(0) = 0   
POL(false) = 0   

The value of delta used in the strict ordering is 4.
The following usable rules [FROCOS05] were oriented:

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

(27) Obligation:

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.

(28) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(29) TRUE