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:

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

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:

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

(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:

MINUS(s(x), s(y)) → MINUS(x, y)
DIV(s(x), s(y)) → DIV(minus(x, y), s(y))
DIV(s(x), s(y)) → MINUS(x, y)
F(x, s(y), b) → DIV(f(x, minus(s(y), s(0)), b), b)
F(x, s(y), b) → F(x, minus(s(y), s(0)), b)
F(x, s(y), b) → MINUS(s(y), s(0))

The TRS R consists of the following rules:

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

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 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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:

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

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:

MINUS(s(x), s(y)) → MINUS(x, y)

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

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

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].

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

(11) Obligation:

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.

(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:

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

(13) TRUE

(14) Obligation:

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

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

The TRS R consists of the following rules:

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

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:

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

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].

div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

(18) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

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

POL(0) = 0   
POL(DIV(x₁, x₂)) = x₁   
POL(minus(x₁, x₂)) = x₁   
POL(s(x₁)) = 1 + x₁   

The following usable rules [FROCOS05] were oriented:

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

(20) Obligation:

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

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

The set Q consists of the following terms:

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

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

(21) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(22) TRUE

(23) Obligation:

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

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

The TRS R consists of the following rules:

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

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

(24) 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.

(25) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

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

(26) 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].

div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

(27) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(28) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule F(x, s(y), b) → F(x, minus(s(y), s(0)), b) at position [1] we obtained the following new rules [LPAR04]:

F(y0, s(0), y2) → F(y0, 0, y2)
F(y0, s(x0), y2) → F(y0, minus(x0, 0), y2)

(29) Obligation:

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

F(y0, s(0), y2) → F(y0, 0, y2)
F(y0, s(x0), y2) → F(y0, minus(x0, 0), y2)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(30) DependencyGraphProof (EQUIVALENT transformation)

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

(31) Obligation:

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

F(y0, s(x0), y2) → F(y0, minus(x0, 0), y2)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(32) 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.

(33) Obligation:

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

F(y0, s(x0), y2) → F(y0, minus(x0, 0), y2)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x

The set Q consists of the following terms:

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

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

(34) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(y0, s(x0), y2) → F(y0, minus(x0, 0), y2) at position [1] we obtained the following new rules [LPAR04]:

F(y0, s(x0), y2) → F(y0, x0, y2)

(35) Obligation:

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

F(y0, s(x0), y2) → F(y0, x0, y2)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x

The set Q consists of the following terms:

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

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

(36) 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.

(37) Obligation:

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

F(y0, s(x0), y2) → F(y0, x0, y2)

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

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

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

(38) 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].

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

(39) Obligation:

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

F(y0, s(x0), y2) → F(y0, x0, y2)

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

(40) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule F(y0, s(x0), y2) → F(y0, x0, y2) we obtained the following new rules [LPAR04]:

F(x0, s(s(y_1)), x2) → F(x0, s(y_1), x2)

(41) Obligation:

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

F(x0, s(s(y_1)), x2) → F(x0, s(y_1), x2)

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

(42) 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:

• F(x0, s(s(y_1)), x2) → F(x0, s(y_1), x2)
  The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3

(43) TRUE