Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
  ↳13 QDP
    ↳14 QDPOrderProof (⇔)
    ↳15 QDP
    ↳16 PisEmptyProof (⇔)
    ↳17 TRUE
  ↳18 QDP
    ↳19 QDPOrderProof (⇔)
    ↳20 QDP
    ↳21 DependencyGraphProof (⇔)
    ↳22 AND
      ↳23 QDP
        ↳24 QDPOrderProof (⇔)
        ↳25 QDP
        ↳26 DependencyGraphProof (⇔)
        ↳27 TRUE
      ↳28 QDP
        ↳29 QDPOrderProof (⇔)
        ↳30 QDP
        ↳31 DependencyGraphProof (⇔)
        ↳32 TRUE

(0) Obligation:

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

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0))))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

PLUS(s(x), y) → PLUS(x, y)
PLUS(s(x), y) → PLUS(p(s(x)), y)
PLUS(s(x), y) → P(s(x))
PLUS(x, s(y)) → PLUS(x, p(s(y)))
PLUS(x, s(y)) → P(s(y))
TIMES(s(x), y) → PLUS(y, times(x, y))
TIMES(s(x), y) → TIMES(x, y)
DIV(x, y) → QUOT(x, y, y)
QUOT(s(x), s(y), z) → QUOT(x, y, z)
QUOT(x, 0, s(z)) → DIV(x, s(z))
DIV(div(x, y), z) → DIV(x, times(zero(y), z))
DIV(div(x, y), z) → TIMES(zero(y), z)
DIV(div(x, y), z) → ZERO(y)
EQ(s(x), s(y)) → EQ(x, y)
DIVIDES(y, x) → EQ(x, times(div(x, y), y))
DIVIDES(y, x) → TIMES(div(x, y), y)
DIVIDES(y, x) → DIV(x, y)
PRIME(s(s(x))) → PR(s(s(x)), s(x))
PR(x, s(s(y))) → IF(divides(s(s(y)), x), x, s(y))
PR(x, s(s(y))) → DIVIDES(s(s(y)), x)
IF(false, x, y) → PR(x, y)
ZERO(s(x)) → IF(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0))))
ZERO(s(x)) → EQ(x, s(0))
ZERO(s(x)) → PLUS(zero(0), 0)
ZERO(s(x)) → ZERO(0)
ZERO(s(x)) → PLUS(0, zero(0))

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0))))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 11 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

EQ(s(x), s(y)) → EQ(x, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0))))

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


EQ(s(x), s(y)) → EQ(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
EQ(x1, x2)  =  x1
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0))))

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

PLUS(s(x), y) → PLUS(p(s(x)), y)
PLUS(s(x), y) → PLUS(x, y)
PLUS(x, s(y)) → PLUS(x, p(s(y)))

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0))))

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PLUS(s(x), y) → PLUS(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2)  =  PLUS(x1)
s(x1)  =  s(x1)
p(x1)  =  x1

Recursive Path Order [RPO].
Precedence:
[PLUS1, s1]


The following usable rules [FROCOS05] were oriented:

p(s(x)) → x

(12) Obligation:

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

PLUS(s(x), y) → PLUS(p(s(x)), y)
PLUS(x, s(y)) → PLUS(x, p(s(y)))

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0))))

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

(13) Obligation:

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

TIMES(s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0))))

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

(14) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TIMES(s(x), y) → TIMES(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TIMES(x1, x2)  =  x1
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(15) Obligation:

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

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0))))

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

(16) PisEmptyProof (EQUIVALENT transformation)

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

(17) TRUE

(18) Obligation:

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

DIV(x, y) → QUOT(x, y, y)
QUOT(s(x), s(y), z) → QUOT(x, y, z)
QUOT(x, 0, s(z)) → DIV(x, s(z))
DIV(div(x, y), z) → DIV(x, times(zero(y), z))
DIV(div(x, y), z) → ZERO(y)
ZERO(s(x)) → IF(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0))))
IF(false, x, y) → PR(x, y)
PR(x, s(s(y))) → IF(divides(s(s(y)), x), x, s(y))
PR(x, s(s(y))) → DIVIDES(s(s(y)), x)
DIVIDES(y, x) → DIV(x, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0))))

Q is empty.
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.


QUOT(s(x), s(y), z) → QUOT(x, y, z)
DIV(div(x, y), z) → DIV(x, times(zero(y), z))
DIV(div(x, y), z) → ZERO(y)
ZERO(s(x)) → IF(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0))))
PR(x, s(s(y))) → DIVIDES(s(s(y)), x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DIV(x1, x2)  =  x1
QUOT(x1, x2, x3)  =  x1
s(x1)  =  s(x1)
0  =  0
div(x1, x2)  =  div(x1, x2)
times(x1, x2)  =  times
zero(x1)  =  zero(x1)
ZERO(x1)  =  x1
IF(x1, x2, x3)  =  IF(x2)
eq(x1, x2)  =  x1
plus(x1, x2)  =  plus(x1)
false  =  false
PR(x1, x2)  =  PR(x1)
divides(x1, x2)  =  divides
DIVIDES(x1, x2)  =  x2
true  =  true
quot(x1, x2, x3)  =  x3
p(x1)  =  x1
if(x1, x2, x3)  =  x2
pr(x1, x2)  =  x2

Recursive Path Order [RPO].
Precedence:
[s1, divides] > zero1 > [0, div2, false] > [times, plus1]
[s1, divides] > [IF1, PR1] > [times, plus1]
[s1, divides] > true > [0, div2, false] > [times, plus1]


The following usable rules [FROCOS05] were oriented:

plus(x, 0) → x

(20) Obligation:

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

DIV(x, y) → QUOT(x, y, y)
QUOT(x, 0, s(z)) → DIV(x, s(z))
IF(false, x, y) → PR(x, y)
PR(x, s(s(y))) → IF(divides(s(s(y)), x), x, s(y))
DIVIDES(y, x) → DIV(x, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0))))

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

(21) DependencyGraphProof (EQUIVALENT transformation)

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

(22) Complex Obligation (AND)

(23) Obligation:

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

PR(x, s(s(y))) → IF(divides(s(s(y)), x), x, s(y))
IF(false, x, y) → PR(x, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0))))

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

(24) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PR(x, s(s(y))) → IF(divides(s(s(y)), x), x, s(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PR(x1, x2)  =  x2
s(x1)  =  s(x1)
IF(x1, x2, x3)  =  x3
divides(x1, x2)  =  divides(x1)
false  =  false
eq(x1, x2)  =  eq
0  =  0
true  =  true
quot(x1, x2, x3)  =  quot(x1, x2, x3)
zero(x1)  =  x1
div(x1, x2)  =  x1
times(x1, x2)  =  times
plus(x1, x2)  =  plus
p(x1)  =  p
if(x1, x2, x3)  =  x3
pr(x1, x2)  =  pr(x1, x2)

Recursive Path Order [RPO].
Precedence:
eq > true > false
quot3 > [s1, divides1, 0]
times > plus > [s1, divides1, 0]
pr2 > [s1, divides1, 0]
pr2 > true > false


The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

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

IF(false, x, y) → PR(x, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0))))

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

(26) DependencyGraphProof (EQUIVALENT transformation)

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

(27) TRUE

(28) Obligation:

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

QUOT(x, 0, s(z)) → DIV(x, s(z))
DIV(x, y) → QUOT(x, y, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0))))

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

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


QUOT(x, 0, s(z)) → DIV(x, s(z))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUOT(x1, x2, x3)  =  QUOT(x1, x2)
0  =  0
s(x1)  =  s
DIV(x1, x2)  =  DIV(x1, x2)

Recursive Path Order [RPO].
Precedence:
0 > [QUOT2, s, DIV2]


The following usable rules [FROCOS05] were oriented: none

(30) Obligation:

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

DIV(x, y) → QUOT(x, y, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(zero(y), s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(zero(y), z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)
zero(div(x, x)) → x
zero(divides(x, x)) → x
zero(times(x, x)) → x
zero(quot(x, x, x)) → x
zero(s(x)) → if(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0))))

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

(31) DependencyGraphProof (EQUIVALENT transformation)

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

(32) TRUE