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(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)

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(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, 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)
PLUS(s(x), y) → PLUS(x, s(y))
DIV(x, y) → QUOT(x, y, 0)
QUOT(x, y, z) → IF(zero(x), x, y, plus(z, s(0)))
QUOT(x, y, z) → ZERO(x)
QUOT(x, y, z) → PLUS(z, s(0))
IF(true, x, y, z) → P(z)
IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
IF(false, x, s(y), z) → MINUS(x, s(y))

The TRS R consists of the following rules:

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, 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 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

PLUS(s(x), y) → PLUS(x, s(y))

The TRS R consists of the following rules:

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, 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:

PLUS(s(x), y) → PLUS(x, s(y))

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

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, 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(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

(11) Obligation:

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

PLUS(s(x), y) → PLUS(x, s(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:

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

(13) TRUE

(14) 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(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, 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:

MINUS(s(x), s(y)) → MINUS(x, y)

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

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, 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].

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

(18) 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.

(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:

• MINUS(s(x), s(y)) → MINUS(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:

IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
QUOT(x, y, z) → IF(zero(x), x, y, plus(z, s(0)))

The TRS R consists of the following rules:

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

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:

IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
QUOT(x, y, z) → IF(zero(x), x, y, plus(z, s(0)))

The TRS R consists of the following rules:

zero(s(x)) → false
zero(0) → true
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

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].

p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

(25) Obligation:

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

IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
QUOT(x, y, z) → IF(zero(x), x, y, plus(z, s(0)))

The TRS R consists of the following rules:

zero(s(x)) → false
zero(0) → true
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x

The set Q consists of the following terms:

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

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

(26) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule QUOT(x, y, z) → IF(zero(x), x, y, plus(z, s(0))) at position [0] we obtained the following new rules [LPAR04]:

QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0)))
QUOT(0, y1, y2) → IF(true, 0, y1, plus(y2, s(0)))

(27) Obligation:

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

IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0)))
QUOT(0, y1, y2) → IF(true, 0, y1, plus(y2, s(0)))

The TRS R consists of the following rules:

zero(s(x)) → false
zero(0) → true
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x

The set Q consists of the following terms:

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

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:

QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0)))
IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)

The TRS R consists of the following rules:

zero(s(x)) → false
zero(0) → true
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x

The set Q consists of the following terms:

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

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

(30) 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.

(31) Obligation:

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

QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0)))
IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(32) 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].

zero(s(x0))
zero(0)

(33) Obligation:

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

QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0)))
IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(34) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z) at position [0] we obtained the following new rules [LPAR04]:

IF(false, 0, s(y1), y2) → QUOT(0, s(y1), y2)
IF(false, s(x0), s(x1), y2) → QUOT(minus(x0, x1), s(x1), y2)

(35) Obligation:

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

QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0)))
IF(false, 0, s(y1), y2) → QUOT(0, s(y1), y2)
IF(false, s(x0), s(x1), y2) → QUOT(minus(x0, x1), s(x1), y2)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(36) DependencyGraphProof (EQUIVALENT transformation)

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

(37) Obligation:

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

IF(false, s(x0), s(x1), y2) → QUOT(minus(x0, x1), s(x1), y2)
QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(38) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0))) we obtained the following new rules [LPAR04]:

QUOT(s(x0), s(z1), z2) → IF(false, s(x0), s(z1), plus(z2, s(0)))

(39) Obligation:

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

IF(false, s(x0), s(x1), y2) → QUOT(minus(x0, x1), s(x1), y2)
QUOT(s(x0), s(z1), z2) → IF(false, s(x0), s(z1), plus(z2, s(0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(40) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(false, s(x0), s(x1), y2) → QUOT(minus(x0, x1), s(x1), y2)

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

POL(0) = 0   
POL(IF(x₁, x₂, x₃, x₄)) = x₁ + x₂   
POL(QUOT(x₁, x₂, x₃)) = 1 + x₁   
POL(false) = 1   
POL(minus(x₁, x₂)) = x₁   
POL(plus(x₁, x₂)) = 0   
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, y) → 0

(41) Obligation:

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

QUOT(s(x0), s(z1), z2) → IF(false, s(x0), s(z1), plus(z2, s(0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(42) DependencyGraphProof (EQUIVALENT transformation)

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

(43) TRUE