Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 UsableRulesProof (⇔)
    ↳7 QDP
    ↳8 QDPSizeChangeProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 UsableRulesProof (⇔)
    ↳12 QDP
    ↳13 MNOCProof (⇔)
    ↳14 QDP
    ↳15 MRRProof (⇔)
    ↳16 QDP
    ↳17 DependencyGraphProof (⇔)
    ↳18 TRUE
  ↳19 QDP
    ↳20 UsableRulesProof (⇔)
    ↳21 QDP
    ↳22 QDPSizeChangeProof (⇔)
    ↳23 TRUE
  ↳24 QDP
    ↳25 QDPOrderProof (⇔)
    ↳26 QDP
    ↳27 DependencyGraphProof (⇔)
    ↳28 AND
      ↳29 QDP
        ↳30 QDPOrderProof (⇔)
        ↳31 QDP
        ↳32 Instantiation (⇔)
        ↳33 QDP
        ↳34 DependencyGraphProof (⇔)
        ↳35 TRUE
      ↳36 QDP
        ↳37 QDPSizeChangeProof (⇔)
        ↳38 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) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

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

EQ(s(x), s(y)) → EQ(x, y)

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

(8) 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:

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

(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) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(12) 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(s(x)) → x

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

(13) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(14) 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(s(x)) → x

The set Q consists of the following terms:

p(s(x0))

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

(15) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

PLUS(s(x), y) → PLUS(x, y)

Strictly oriented rules of the TRS R:

p(s(x)) → x

Used ordering: Polynomial interpretation [POLO]:

POL(PLUS(x1, x2)) = x1 + x2   
POL(p(x1)) = x1   
POL(s(x1)) = 1 + x1   

(16) 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)))

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

p(s(x0))

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

(17) DependencyGraphProof (EQUIVALENT transformation)

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

(18) TRUE

(19) 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.

(20) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(21) Obligation:

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

TIMES(s(x), y) → TIMES(x, y)

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

(22) 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:

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

(23) TRUE

(24) 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.

(25) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


DIV(div(x, y), z) → DIV(x, times(zero(y), z))
ZERO(s(x)) → IF(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(DIV(x1, x2)) = x1   
POL(DIVIDES(x1, x2)) = x2   
POL(IF(x1, x2, x3)) = x2   
POL(PR(x1, x2)) = x1   
POL(QUOT(x1, x2, x3)) = x1   
POL(ZERO(x1)) = 1   
POL(div(x1, x2)) = 1 + x1   
POL(divides(x1, x2)) = 0   
POL(eq(x1, x2)) = 0   
POL(false) = 0   
POL(if(x1, x2, x3)) = 0   
POL(p(x1)) = 0   
POL(plus(x1, x2)) = x1   
POL(pr(x1, x2)) = 0   
POL(quot(x1, x2, x3)) = 0   
POL(s(x1)) = x1   
POL(times(x1, x2)) = 0   
POL(true) = 0   
POL(zero(x1)) = 0   

The following usable rules [FROCOS05] were oriented:

plus(x, 0) → x

(26) 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) → ZERO(y)
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.

(27) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.

(28) Complex Obligation (AND)

(29) Obligation:

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

QUOT(s(x), s(y), z) → QUOT(x, y, z)
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.

(30) 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)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(DIV(x1, x2)) = x1   
POL(QUOT(x1, x2, x3)) = x1   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(31) 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.

(32) Instantiation (EQUIVALENT transformation)

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

DIV(y_0, s(y_1)) → QUOT(y_0, s(y_1), s(y_1))

(33) Obligation:

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

QUOT(x, 0, s(z)) → DIV(x, s(z))
DIV(y_0, s(y_1)) → QUOT(y_0, s(y_1), s(y_1))

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.

(34) DependencyGraphProof (EQUIVALENT transformation)

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

(35) TRUE

(36) 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.

(37) 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:

  • IF(false, x, y) → PR(x, y)
    The graph contains the following edges 2 >= 1, 3 >= 2

  • PR(x, s(s(y))) → IF(divides(s(s(y)), x), x, s(y))
    The graph contains the following edges 1 >= 2, 2 > 3

(38) TRUE