11/06/2011, 08:06 PM
I realize now we have to create a second axiom in order that the series of operators become the true Ackermann function.
Consider the possibility that:
\( a\,\,\bigtriangleup_2\,\,2=a \)
we still have the result that
\( a + (a\,\,\bigtriangleup_2\,\,2) = a\,\, \bigtriangleup_2\,\, (2 + 1) = a\,\,\bigtriangleup_2\,\,3 \)
the only difference is that
\( S(2) = 2 \) and in return we get
\( a\,\,\bigtriangleup_2\,\,b = a \cdot (b-1) \)
So in order to get the true Ackermann function we must make the second assertion:
\( n\,\,\in\,\,\mathbb{Z};n \ge 2 \)
\( S(n) = 1 \)
This is actually the equivalent to the iteration axiom:
\( a\,\,\bigtriangleup_{\sigma + 1}\,\,n = a_1\,\,\bigtriangleup_\sigma\,\,(a_2\,\,\bigtriangleup_\sigma\,\,(a_3\,\, \bigtriangleup_\sigma \,\,...(a_{n-1}\,\,\bigtriangleup_\sigma\,\,a_n) \)
though only true for integer values of sigma greater than or equal to two.
Therefore we define the Ackermann function from axioms:
\( \vartheta(a,b,\sigma) = a\,\,\bigtriangleup_\sigma\,\,b \)
such that
\( a\,\,\bigtriangleup_\sigma\,\,(a\,\,\bigtriangleup_{\sigma + 1}\,\,b) = a\,\,\bigtriangleup_\sigma\,\,(b+1) \)
\( a\,\,\bigtriangleup_\sigma\,\,S(\sigma) = a \)
\( n\,\,\in\,\,\mathbb{Z}\,\,;\,\,n \ge 2 \)
\( S(n) = 1 \)
\( a\,\,\bigtriangleup_1\,\,b = a + b \)
Does anyone see any modifications necessary?
Consider the possibility that:
\( a\,\,\bigtriangleup_2\,\,2=a \)
we still have the result that
\( a + (a\,\,\bigtriangleup_2\,\,2) = a\,\, \bigtriangleup_2\,\, (2 + 1) = a\,\,\bigtriangleup_2\,\,3 \)
the only difference is that
\( S(2) = 2 \) and in return we get
\( a\,\,\bigtriangleup_2\,\,b = a \cdot (b-1) \)
So in order to get the true Ackermann function we must make the second assertion:
\( n\,\,\in\,\,\mathbb{Z};n \ge 2 \)
\( S(n) = 1 \)
This is actually the equivalent to the iteration axiom:
\( a\,\,\bigtriangleup_{\sigma + 1}\,\,n = a_1\,\,\bigtriangleup_\sigma\,\,(a_2\,\,\bigtriangleup_\sigma\,\,(a_3\,\, \bigtriangleup_\sigma \,\,...(a_{n-1}\,\,\bigtriangleup_\sigma\,\,a_n) \)
though only true for integer values of sigma greater than or equal to two.
Therefore we define the Ackermann function from axioms:
\( \vartheta(a,b,\sigma) = a\,\,\bigtriangleup_\sigma\,\,b \)
such that
\( a\,\,\bigtriangleup_\sigma\,\,(a\,\,\bigtriangleup_{\sigma + 1}\,\,b) = a\,\,\bigtriangleup_\sigma\,\,(b+1) \)
\( a\,\,\bigtriangleup_\sigma\,\,S(\sigma) = a \)
\( n\,\,\in\,\,\mathbb{Z}\,\,;\,\,n \ge 2 \)
\( S(n) = 1 \)
\( a\,\,\bigtriangleup_1\,\,b = a + b \)
Does anyone see any modifications necessary?