If $T(n)$ implies $T(2n)$, $T(n)$ implies $T(n-5)$ for $n \geq 6$ and $T(1)$ is true, then is $T(n)$ true for all $n$ in natural numbers? My reasoning:
From the assumption it follows that for all numbers in form $n = 2^m - 5k$, where $n,m,k$ are some natural numbers  $T(n)$ is true. When $n=5$ then $(5 = 2^m - 5k) \equiv 5(k+1) = 2^m$, therefore $n$ can't equal $5$. Hence we can't conclude (from given assumptions) that $T(n)$ is true for all $n$ in natural numbers. Neither we can say that $T(n)$ is not true for all natural numbers. So my answer would be that we can't tell. Am I wrong?
 A: For the edited version of the question: no, you can't conclude that $T(n)$ is true for every $n$.
Perhaps the easiest way to show this is to notice that if $T(n)$ is the proposition "$n$ is not a multiple of $5$" then it satisfies all three requirements, but is not true for all natural numbers.
A: Hint: consider residue classes modulo $5$ - what can you say about these?
A: Hint:  If $k \equiv 1 \pmod 5$ then $2k \equiv 2 \equiv k + 1 \mod 5$.
If $k\equiv 2 \pmod 5$ then $4k \equiv 8 \equiv 3 \equiv k+1 \pmod 5$.
If $k \equiv 3\pmod 5$ then $8k \equiv 24 \equiv  4 \equiv k+ 1\mod 5$.
if $k \equiv 4 \pmod 5$ then .... dang.
Okay, that's it.  You can't get any multiple of $5$.
If $k= 5m$ and we able to conclude that $T(k)$ were true could only do it by assuming it is true for some $n$ were $k = 2n$ or $k = n-5$.  In either case $n$ must be a multiple of $5$ and we are boot strapping.
Hmmm, I'm not sure how to formally order thn $n$ so the we can use a well ordering principal to say there we could deduce all natural numbers there must be a "first" multiple of $5$  and that's contradictory (as there is no order of deduction that we "must" follow). But it is clear we can not deduce a multiple of $5$ without first deducing some other multiple of $5$ first.
We can deduce all other numbers though.  (If $k \equiv  4 \pmod 5$ then $4k \equiv 1\equiv k+2 \pmod 5$.)
