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:

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

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:

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(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:

GE(s(x), s(y)) → GE(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
DIV(x, y) → IFY(ge(y, s(0)), x, y)
DIV(x, y) → GE(y, s(0))
IFY(true, x, y) → IF(ge(x, y), x, y)
IFY(true, x, y) → GE(x, y)
IF(true, x, y) → DIV(minus(x, y), y)
IF(true, x, y) → MINUS(x, y)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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:

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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].

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(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:

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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:

GE(s(x), s(y)) → GE(x, y)

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

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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].

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(18) Obligation:

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

GE(s(x), s(y)) → GE(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:

• GE(s(x), s(y)) → GE(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:

DIV(x, y) → IFY(ge(y, s(0)), x, y)
IFY(true, x, y) → IF(ge(x, y), x, y)
IF(true, x, y) → DIV(minus(x, y), y)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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:

DIV(x, y) → IFY(ge(y, s(0)), x, y)
IFY(true, x, y) → IF(ge(x, y), x, y)
IF(true, x, y) → DIV(minus(x, y), y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
minus(x0, 0)
minus(s(x0), s(x1))
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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].

div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(25) Obligation:

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

DIV(x, y) → IFY(ge(y, s(0)), x, y)
IFY(true, x, y) → IF(ge(x, y), x, y)
IF(true, x, y) → DIV(minus(x, y), y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

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

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

(26) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule DIV(x, y) → IFY(ge(y, s(0)), x, y) at position [0] we obtained the following new rules [LPAR04]:

DIV(y0, 0) → IFY(false, y0, 0)
DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))

(27) Obligation:

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

IFY(true, x, y) → IF(ge(x, y), x, y)
IF(true, x, y) → DIV(minus(x, y), y)
DIV(y0, 0) → IFY(false, y0, 0)
DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

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

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

(28) DependencyGraphProof (EQUIVALENT transformation)

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

(29) Obligation:

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

IF(true, x, y) → DIV(minus(x, y), y)
DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))
IFY(true, x, y) → IF(ge(x, y), x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

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

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

(30) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0)) at position [0] we obtained the following new rules [LPAR04]:

DIV(y0, s(x0)) → IFY(true, y0, s(x0))

(31) Obligation:

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

IF(true, x, y) → DIV(minus(x, y), y)
IFY(true, x, y) → IF(ge(x, y), x, y)
DIV(y0, s(x0)) → IFY(true, y0, s(x0))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

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

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

(32) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IFY(true, x, y) → IF(ge(x, y), x, y) at position [0] we obtained the following new rules [LPAR04]:

IFY(true, x0, 0) → IF(true, x0, 0)
IFY(true, 0, s(x0)) → IF(false, 0, s(x0))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))

(33) Obligation:

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

IF(true, x, y) → DIV(minus(x, y), y)
DIV(y0, s(x0)) → IFY(true, y0, s(x0))
IFY(true, x0, 0) → IF(true, x0, 0)
IFY(true, 0, s(x0)) → IF(false, 0, s(x0))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

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

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

(34) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(35) Obligation:

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

DIV(y0, s(x0)) → IFY(true, y0, s(x0))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, x, y) → DIV(minus(x, y), y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

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

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

(36) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(true, x, y) → DIV(minus(x, y), y) we obtained the following new rules [LPAR04]:

IF(true, s(z0), s(z1)) → DIV(minus(s(z0), s(z1)), s(z1))

(37) Obligation:

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

DIV(y0, s(x0)) → IFY(true, y0, s(x0))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, s(z0), s(z1)) → DIV(minus(s(z0), s(z1)), s(z1))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

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

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

(38) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF(true, s(z0), s(z1)) → DIV(minus(s(z0), s(z1)), s(z1)) at position [0] we obtained the following new rules [LPAR04]:

IF(true, s(z0), s(z1)) → DIV(minus(z0, z1), s(z1))

(39) Obligation:

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

DIV(y0, s(x0)) → IFY(true, y0, s(x0))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, s(z0), s(z1)) → DIV(minus(z0, z1), s(z1))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

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

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

(40) ForwardInstantiation (EQUIVALENT transformation)

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

DIV(s(y_0), s(x1)) → IFY(true, s(y_0), s(x1))

(41) Obligation:

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

IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, s(z0), s(z1)) → DIV(minus(z0, z1), s(z1))
DIV(s(y_0), s(x1)) → IFY(true, s(y_0), s(x1))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

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

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

(42) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IF(true, s(z0), s(z1)) → DIV(minus(z0, z1), s(z1)) at position [0] we obtained the following new rules [LPAR04]:

IF(true, s(x0), s(0)) → DIV(x0, s(0))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))

(43) Obligation:

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

IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
DIV(s(y_0), s(x1)) → IFY(true, s(y_0), s(x1))
IF(true, s(x0), s(0)) → DIV(x0, s(0))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

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

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

(44) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule DIV(s(y_0), s(x1)) → IFY(true, s(y_0), s(x1)) we obtained the following new rules [LPAR04]:

DIV(s(x0), s(0)) → IFY(true, s(x0), s(0))
DIV(s(x0), s(s(z1))) → IFY(true, s(x0), s(s(z1)))

(45) Obligation:

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

IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, s(x0), s(0)) → DIV(x0, s(0))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
DIV(s(x0), s(0)) → IFY(true, s(x0), s(0))
DIV(s(x0), s(s(z1))) → IFY(true, s(x0), s(s(z1)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

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

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

(46) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1)) we obtained the following new rules [LPAR04]:

IFY(true, s(z0), s(0)) → IF(ge(z0, 0), s(z0), s(0))
IFY(true, s(z0), s(s(z1))) → IF(ge(z0, s(z1)), s(z0), s(s(z1)))

(47) Obligation:

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

IF(true, s(x0), s(0)) → DIV(x0, s(0))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
DIV(s(x0), s(0)) → IFY(true, s(x0), s(0))
DIV(s(x0), s(s(z1))) → IFY(true, s(x0), s(s(z1)))
IFY(true, s(z0), s(0)) → IF(ge(z0, 0), s(z0), s(0))
IFY(true, s(z0), s(s(z1))) → IF(ge(z0, s(z1)), s(z0), s(s(z1)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

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

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

(48) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.

(49) Complex Obligation (AND)

(50) Obligation:

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

DIV(s(x0), s(s(z1))) → IFY(true, s(x0), s(s(z1)))
IFY(true, s(z0), s(s(z1))) → IF(ge(z0, s(z1)), s(z0), s(s(z1)))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

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

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

(51) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IFY(true, s(z0), s(s(z1))) → IF(ge(z0, s(z1)), s(z0), s(s(z1))) we obtained the following new rules [LPAR04]:

IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(s(y_1), s(x1)), s(s(y_1)), s(s(x1)))

(52) Obligation:

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

DIV(s(x0), s(s(z1))) → IFY(true, s(x0), s(s(z1)))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(s(y_1), s(x1)), s(s(y_1)), s(s(x1)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

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

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

(53) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(s(y_1), s(x1)), s(s(y_1)), s(s(x1))) at position [0] we obtained the following new rules [LPAR04]:

IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(y_1, x1), s(s(y_1)), s(s(x1)))

(54) Obligation:

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

DIV(s(x0), s(s(z1))) → IFY(true, s(x0), s(s(z1)))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(y_1, x1), s(s(y_1)), s(s(x1)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

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

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

(55) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))

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

POL(0) = 0   
POL(DIV(x₁, x₂)) = x₁   
POL(IF(x₁, x₂, x₃)) = x₂   
POL(IFY(x₁, x₂, x₃)) = x₂   
POL(false) = 0   
POL(ge(x₁, x₂)) = 0   
POL(minus(x₁, x₂)) = x₁   
POL(s(x₁)) = 1 + x₁   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented:

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

(56) Obligation:

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

DIV(s(x0), s(s(z1))) → IFY(true, s(x0), s(s(z1)))
IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(y_1, x1), s(s(y_1)), s(s(x1)))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

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

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

(57) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(58) TRUE

(59) Obligation:

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

DIV(s(x0), s(0)) → IFY(true, s(x0), s(0))
IFY(true, s(z0), s(0)) → IF(ge(z0, 0), s(z0), s(0))
IF(true, s(x0), s(0)) → DIV(x0, s(0))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

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

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

(60) 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.

(61) Obligation:

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

DIV(s(x0), s(0)) → IFY(true, s(x0), s(0))
IFY(true, s(z0), s(0)) → IF(ge(z0, 0), s(z0), s(0))
IF(true, s(x0), s(0)) → DIV(x0, s(0))

The TRS R consists of the following rules:

ge(x, 0) → true

The set Q consists of the following terms:

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

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

(62) 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, 0)
minus(s(x0), s(x1))

(63) Obligation:

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

DIV(s(x0), s(0)) → IFY(true, s(x0), s(0))
IFY(true, s(z0), s(0)) → IF(ge(z0, 0), s(z0), s(0))
IF(true, s(x0), s(0)) → DIV(x0, s(0))

The TRS R consists of the following rules:

ge(x, 0) → true

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(64) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IFY(true, s(z0), s(0)) → IF(ge(z0, 0), s(z0), s(0)) at position [0] we obtained the following new rules [LPAR04]:

IFY(true, s(z0), s(0)) → IF(true, s(z0), s(0))

(65) Obligation:

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

DIV(s(x0), s(0)) → IFY(true, s(x0), s(0))
IF(true, s(x0), s(0)) → DIV(x0, s(0))
IFY(true, s(z0), s(0)) → IF(true, s(z0), s(0))

The TRS R consists of the following rules:

ge(x, 0) → true

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(66) 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.

(67) Obligation:

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

DIV(s(x0), s(0)) → IFY(true, s(x0), s(0))
IF(true, s(x0), s(0)) → DIV(x0, s(0))
IFY(true, s(z0), s(0)) → IF(true, s(z0), s(0))

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

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(68) 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].

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

(69) Obligation:

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

DIV(s(x0), s(0)) → IFY(true, s(x0), s(0))
IF(true, s(x0), s(0)) → DIV(x0, s(0))
IFY(true, s(z0), s(0)) → IF(true, s(z0), s(0))

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

(70) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IF(true, s(x0), s(0)) → DIV(x0, s(0)) we obtained the following new rules [LPAR04]:

IF(true, s(s(y_0)), s(0)) → DIV(s(y_0), s(0))

(71) Obligation:

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

DIV(s(x0), s(0)) → IFY(true, s(x0), s(0))
IFY(true, s(z0), s(0)) → IF(true, s(z0), s(0))
IF(true, s(s(y_0)), s(0)) → DIV(s(y_0), s(0))

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

(72) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IFY(true, s(z0), s(0)) → IF(true, s(z0), s(0)) we obtained the following new rules [LPAR04]:

IFY(true, s(s(y_0)), s(0)) → IF(true, s(s(y_0)), s(0))

(73) Obligation:

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

DIV(s(x0), s(0)) → IFY(true, s(x0), s(0))
IF(true, s(s(y_0)), s(0)) → DIV(s(y_0), s(0))
IFY(true, s(s(y_0)), s(0)) → IF(true, s(s(y_0)), s(0))

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

(74) 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:

• IFY(true, s(s(y_0)), s(0)) → IF(true, s(s(y_0)), s(0))
  The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3

• IF(true, s(s(y_0)), s(0)) → DIV(s(y_0), s(0))
  The graph contains the following edges 2 > 1, 3 >= 2

• DIV(s(x0), s(0)) → IFY(true, s(x0), s(0))
  The graph contains the following edges 1 >= 2, 2 >= 3

(75) TRUE