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:

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(a) = 0   
POL(b) = 0   
POL(double(x₁)) = x₁   
POL(exp(x₁)) = x₁   
POL(f(x₁, x₂, x₃)) = 1 + x₁ + x₂ + x₃   
POL(half(x₁)) = x₁   
POL(s(x₁)) = x₁   
POL(tower(x₁)) = 1 + x₁   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

f(a, 0, y) → y

(2) Obligation:

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

tower(x) → f(a, x, s(0))
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(a) = 0   
POL(b) = 0   
POL(double(x₁)) = x₁   
POL(exp(x₁)) = x₁   
POL(f(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(half(x₁)) = x₁   
POL(s(x₁)) = x₁   
POL(tower(x₁)) = 1 + x₁   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

tower(x) → f(a, x, s(0))

(4) Obligation:

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

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

Q is empty.

(5) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The TRS R 2 is

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))

The signature Sigma is {f}

(6) Obligation:

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

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

(7) DependencyPairsProof (EQUIVALENT transformation)

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

(8) Obligation:

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

F(a, s(x), y) → F(b, y, s(x))
F(b, y, x) → F(a, half(x), exp(y))
F(b, y, x) → HALF(x)
F(b, y, x) → EXP(y)
EXP(s(x)) → DOUBLE(exp(x))
EXP(s(x)) → EXP(x)
DOUBLE(s(x)) → DOUBLE(x)
HALF(0) → DOUBLE(0)
HALF(s(0)) → HALF(0)
HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(9) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 5 less nodes.

(10) Complex Obligation (AND)

(11) Obligation:

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

DOUBLE(s(x)) → DOUBLE(x)

The TRS R consists of the following rules:

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(12) 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.

(13) Obligation:

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

DOUBLE(s(x)) → DOUBLE(x)

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

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(14) 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].

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

(15) Obligation:

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

DOUBLE(s(x)) → DOUBLE(x)

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

(16) 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:

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

(17) TRUE

(18) 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:

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(19) 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.

(20) 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:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(21) 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].

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

(22) 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.

(23) 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

(24) TRUE

(25) Obligation:

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

EXP(s(x)) → EXP(x)

The TRS R consists of the following rules:

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(26) 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.

(27) Obligation:

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

EXP(s(x)) → EXP(x)

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

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(28) 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].

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

(29) Obligation:

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

EXP(s(x)) → EXP(x)

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

(30) 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:

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

(31) TRUE

(32) Obligation:

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

F(b, y, x) → F(a, half(x), exp(y))
F(a, s(x), y) → F(b, y, s(x))

The TRS R consists of the following rules:

f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(33) 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.

(34) Obligation:

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

F(b, y, x) → F(a, half(x), exp(y))
F(a, s(x), y) → F(b, y, s(x))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(35) 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].

f(a, s(x0), x1)
f(b, x0, x1)

(36) Obligation:

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

F(b, y, x) → F(a, half(x), exp(y))
F(a, s(x), y) → F(b, y, s(x))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(37) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(b, y, x) → F(a, half(x), exp(y)) we obtained the following new rules [LPAR04]:

F(b, z1, s(z0)) → F(a, half(s(z0)), exp(z1))

(38) Obligation:

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

F(a, s(x), y) → F(b, y, s(x))
F(b, z1, s(z0)) → F(a, half(s(z0)), exp(z1))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(39) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule F(b, y, x) → F(a, half(x), exp(y)) at position [1] we obtained the following new rules [LPAR04]:

F(b, y0, 0) → F(a, double(0), exp(y0))
F(b, y0, s(0)) → F(a, half(0), exp(y0))
F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))

(40) Obligation:

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

F(a, s(x), y) → F(b, y, s(x))
F(b, y0, 0) → F(a, double(0), exp(y0))
F(b, y0, s(0)) → F(a, half(0), exp(y0))
F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(41) DependencyGraphProof (EQUIVALENT transformation)

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

(42) Obligation:

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

F(b, y0, s(0)) → F(a, half(0), exp(y0))
F(a, s(x), y) → F(b, y, s(x))
F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(43) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(b, y0, s(0)) → F(a, half(0), exp(y0)) at position [1] we obtained the following new rules [LPAR04]:

F(b, y0, s(0)) → F(a, double(0), exp(y0))

(44) Obligation:

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

F(a, s(x), y) → F(b, y, s(x))
F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))
F(b, y0, s(0)) → F(a, double(0), exp(y0))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(45) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(b, y0, s(0)) → F(a, double(0), exp(y0)) at position [1] we obtained the following new rules [LPAR04]:

F(b, y0, s(0)) → F(a, 0, exp(y0))

(46) Obligation:

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

F(a, s(x), y) → F(b, y, s(x))
F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))
F(b, y0, s(0)) → F(a, 0, exp(y0))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(47) DependencyGraphProof (EQUIVALENT transformation)

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

(48) Obligation:

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

F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))
F(a, s(x), y) → F(b, y, s(x))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(49) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule F(a, s(x), y) → F(b, y, s(x)) we obtained the following new rules [LPAR04]:

F(a, s(s(y_1)), x1) → F(b, x1, s(s(y_1)))

(50) Obligation:

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

F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))
F(a, s(s(y_1)), x1) → F(b, x1, s(s(y_1)))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

The set Q consists of the following terms:

exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

(51) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.

(52) Obligation:

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

F(b, y0, s(s(x0))) → F(a, s(half(x0)), exp(y0))
F(a, s(s(y_1)), x1) → F(b, x1, s(s(y_1)))

The TRS R consists of the following rules:

half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))

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

(53) SemLabProof (SOUND transformation)

We found the following model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.a: 0
half: 0
b: 1
s: 0
double: 0
0: 0
F: 0
exp: 0
By semantic labelling [SEMLAB] we obtain the following labelled TRS.

(54) Obligation:

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

F.1-0-0(b., y0, s.0(s.0(x0))) → F.0-0-0(a., s.0(half.0(x0)), exp.0(y0))
F.0-0-0(a., s.0(s.0(y_1)), x1) → F.1-0-0(b., x1, s.0(s.0(y_1)))
F.0-0-1(a., s.0(s.0(y_1)), x1) → F.1-1-0(b., x1, s.0(s.0(y_1)))
F.0-0-0(a., s.0(s.1(y_1)), x1) → F.1-0-0(b., x1, s.0(s.1(y_1)))
F.0-0-1(a., s.0(s.1(y_1)), x1) → F.1-1-0(b., x1, s.0(s.1(y_1)))
F.1-0-0(b., y0, s.0(s.1(x0))) → F.0-0-0(a., s.0(half.1(x0)), exp.0(y0))
F.1-1-0(b., y0, s.0(s.0(x0))) → F.0-0-0(a., s.0(half.0(x0)), exp.1(y0))
F.1-1-0(b., y0, s.0(s.1(x0))) → F.0-0-0(a., s.0(half.1(x0)), exp.1(y0))

The TRS R consists of the following rules:

double.0(s.1(x)) → s.0(s.0(double.1(x)))
half.0(0.) → double.0(0.)
half.0(s.0(0.)) → half.0(0.)
half.0(s.0(s.0(x))) → s.0(half.0(x))
double.0(0.) → 0.
double.0(s.0(x)) → s.0(s.0(double.0(x)))
half.0(s.0(s.1(x))) → s.0(half.1(x))
exp.0(s.1(x)) → double.0(exp.1(x))
exp.0(0.) → s.0(0.)
exp.0(s.0(x)) → double.0(exp.0(x))

The set Q consists of the following terms:

exp.0(0.)
exp.0(s.0(x0))
exp.0(s.1(x0))
double.0(0.)
double.0(s.0(x0))
double.0(s.1(x0))
half.0(0.)
half.0(s.0(0.))
half.0(s.0(s.0(x0)))
half.0(s.0(s.1(x0)))

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

(55) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 6 less nodes.

(56) Obligation:

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

F.0-0-0(a., s.0(s.0(y_1)), x1) → F.1-0-0(b., x1, s.0(s.0(y_1)))
F.1-0-0(b., y0, s.0(s.0(x0))) → F.0-0-0(a., s.0(half.0(x0)), exp.0(y0))

The TRS R consists of the following rules:

double.0(s.1(x)) → s.0(s.0(double.1(x)))
half.0(0.) → double.0(0.)
half.0(s.0(0.)) → half.0(0.)
half.0(s.0(s.0(x))) → s.0(half.0(x))
double.0(0.) → 0.
double.0(s.0(x)) → s.0(s.0(double.0(x)))
half.0(s.0(s.1(x))) → s.0(half.1(x))
exp.0(s.1(x)) → double.0(exp.1(x))
exp.0(0.) → s.0(0.)
exp.0(s.0(x)) → double.0(exp.0(x))

The set Q consists of the following terms:

exp.0(0.)
exp.0(s.0(x0))
exp.0(s.1(x0))
double.0(0.)
double.0(s.0(x0))
double.0(s.1(x0))
half.0(0.)
half.0(s.0(0.))
half.0(s.0(s.0(x0)))
half.0(s.0(s.1(x0)))

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

(57) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

half.0(s.0(s.1(x))) → s.0(half.1(x))
exp.0(s.1(x)) → double.0(exp.1(x))
double.0(s.1(x)) → s.0(s.0(double.1(x)))

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(0.) = 0   
POL(F.0-0-0(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(F.1-0-0(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(a.) = 0   
POL(b.) = 0   
POL(double.0(x₁)) = x₁   
POL(double.1(x₁)) = x₁   
POL(exp.0(x₁)) = x₁   
POL(exp.1(x₁)) = x₁   
POL(half.0(x₁)) = x₁   
POL(half.1(x₁)) = x₁   
POL(s.0(x₁)) = x₁   
POL(s.1(x₁)) = x₁   

(58) Obligation:

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

F.0-0-0(a., s.0(s.0(y_1)), x1) → F.1-0-0(b., x1, s.0(s.0(y_1)))
F.1-0-0(b., y0, s.0(s.0(x0))) → F.0-0-0(a., s.0(half.0(x0)), exp.0(y0))

The TRS R consists of the following rules:

half.0(0.) → double.0(0.)
half.0(s.0(0.)) → half.0(0.)
half.0(s.0(s.0(x))) → s.0(half.0(x))
exp.0(0.) → s.0(0.)
exp.0(s.0(x)) → double.0(exp.0(x))
double.0(0.) → 0.
double.0(s.0(x)) → s.0(s.0(double.0(x)))

The set Q consists of the following terms:

exp.0(0.)
exp.0(s.0(x0))
exp.0(s.1(x0))
double.0(0.)
double.0(s.0(x0))
double.0(s.1(x0))
half.0(0.)
half.0(s.0(0.))
half.0(s.0(s.0(x0)))
half.0(s.0(s.1(x0)))

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

(59) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F.1-0-0(b., y0, s.0(s.0(x0))) → F.0-0-0(a., s.0(half.0(x0)), exp.0(y0))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0.) = 0   
POL(F.0-0-0(x₁, x₂, x₃)) = 1 + x₁ + x₂   
POL(F.1-0-0(x₁, x₂, x₃)) = 1 + x₁ + x₃   
POL(a.) = 1   
POL(b.) = 1   
POL(double.0(x₁)) = 0   
POL(exp.0(x₁)) = 0   
POL(half.0(x₁)) = x₁   
POL(s.0(x₁)) = 1 + x₁   

The following usable rules [FROCOS05] were oriented:

double.0(0.) → 0.
half.0(s.0(s.0(x))) → s.0(half.0(x))
half.0(s.0(0.)) → half.0(0.)
half.0(0.) → double.0(0.)

(60) Obligation:

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

F.0-0-0(a., s.0(s.0(y_1)), x1) → F.1-0-0(b., x1, s.0(s.0(y_1)))

The TRS R consists of the following rules:

half.0(0.) → double.0(0.)
half.0(s.0(0.)) → half.0(0.)
half.0(s.0(s.0(x))) → s.0(half.0(x))
exp.0(0.) → s.0(0.)
exp.0(s.0(x)) → double.0(exp.0(x))
double.0(0.) → 0.
double.0(s.0(x)) → s.0(s.0(double.0(x)))

The set Q consists of the following terms:

exp.0(0.)
exp.0(s.0(x0))
exp.0(s.1(x0))
double.0(0.)
double.0(s.0(x0))
double.0(s.1(x0))
half.0(0.)
half.0(s.0(0.))
half.0(s.0(s.0(x0)))
half.0(s.0(s.1(x0)))

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

(61) DependencyGraphProof (EQUIVALENT transformation)

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

(62) TRUE