Term Rewriting System R:
[x, y, z]
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)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(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)

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

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)
DIV(div(x, y), z) -> DIV(x, times(y, z))
DIV(div(x, y), z) -> TIMES(y, z)
QUOT(s(x), s(y), z) -> QUOT(x, y, z)
QUOT(x, 0, s(z)) -> DIV(x, s(z))
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)

Furthermore, R contains five SCCs.


   R
DPs
       →DP Problem 1
Argument Filtering and Ordering
       →DP Problem 2
Remaining
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pairs:

PLUS(x, s(y)) -> PLUS(x, p(s(y)))
PLUS(s(x), y) -> PLUS(p(s(x)), y)
PLUS(s(x), y) -> PLUS(x, y)


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)
div(div(x, y), z) -> div(x, times(y, z))
quot(0, s(y), z) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(div(x, s(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)


Strategy:

innermost




The following dependency pair can be strictly oriented:

PLUS(s(x), y) -> PLUS(x, y)


The following usable rules for innermost w.r.t. to the AFS can be oriented:

p(0) -> 0
p(s(x)) -> x


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
PLUS(x1, x2) -> PLUS(x1, x2)
s(x1) -> s(x1)
p(x1) -> x1


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Remaining Obligation(s)
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Remaining Obligation(s)
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Remaining Obligation(s)
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Remaining Obligation(s)
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Remaining Obligation(s)
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
Remaining Obligation(s)




The following remains to be proven:

Innermost Termination of R could not be shown.
Duration:
0:00 minutes