Equality by iteratively applying $(a,b)\rightarrow [(a+1,2b)\text{ or }(2a,b+1)]$?

Call a pair equalizable if there is a procedure ending in equal numbers. Clearly any pair of the form $(k,4k)$ is equalizable: $$ (k,4k)\to (2k,4k+1)\to(2k+1,8k+2)\to(4k+2,8k+3)\to(8k+4,8k+4). $$ From this we deduce that any pair $(k,2k)$ is equalizable: $$ (k,2k)\to (k+1,4k)\to(2k+2,4k+1)\to(4k+4,4k+2)\to(8k+8,4k+3)\to (16k+16,4k+4). $$ Hence any pair that has a procedure leading to $(m,m-1)$ or $(m,m-2)$, is equalizable; take $(m,m-1)\to (2m,m)$ and $(m,m-2)\to (2m,m-1)\to (4m,m)$.

From this we deduce that any pair $(k,8k)$ is also equalizable: $$ (k,8k)\to (4k,8k+2)\to(4k+1,16k+4)\to(16k+4,16k+6). $$ Since the difference is 2, it is equalizable.

Hence, by the same argument as above, any pair that has a procedure leading to $(m,m-3)$, is equalizable.

Now we prove that any pair $(k,16k)$ is also equalizable: $$ (k,16k)\to (4k,16k+2)\to(4k+1,32k+4)\to(32k+8,32k+7). $$ Since the difference is 1, it is equalizable. Hence, by the same argument as above, any pair that has a procedure leading to $(m,m-4)$, is equalizable.

Now we prove that any pair $(k,32k)$ is also equalizable: $$ (k,32k)\to (8k,32k+3)\to(8k+1,64k+6)\to(64k+8,64k+9). $$ Since the difference is 1, it is equalizable. Hence, by the same argument as above, any pair that has a procedure leading to $(m,m-5)$, is equalizable.

Now we proceed by induction: Assume that we have proved that any pair of the form $(k,k \cdot 2^j)$ is equalizable, for $j=1,\dots,n-1$. Then it follows as above, that any pair $(s,t)$ with $|s-t|<n$ is equalizable. Below we will prove that for any $n>5$, there exists $r$ with $1<r<n$ such that $$ |2^{n+1-r}-(n+r+1)|<n. $$

Now we prove that $(k,k\cdot 2^n)$ is equalizable: $$ (k,k\cdot 2^n)\to (k\cdot 2^r,k\cdot 2^n+r)\to (k\cdot 2^r+1,k\cdot 2^{n+1}+2r)\to $$ $$ \small \to(k\cdot 2^{n+1}+2^{n+1-r},k\cdot 2^{n+1}+2r+(n+1-r))=(k\cdot 2^{n+1}+2^{n+1-r},k\cdot 2^{n+1}+n+1+r) $$ For the difference we have $|2^{n+1-r}-(n+r+1)|<n$, hence the pair $(k,k\cdot 2^n)$ is equalizable.

Finally we prove that for any $n>5$, there exists $r$ with $1<r<n$ such that $$ |2^{n+1-r}-(n+r+1)|<n. $$ We define the decreasing function $f(k):=2^{n+1-k}-(n+k+1)$ which fulfills $f(1)>0$ and $f(n-2)=8-(2n-1)<0$. Hence there exists $k\le n-3$ with $f(k)\ge 0$ and $f(k+1)<0$. Then $$ |f(k)|+|f(k+1)|=2^{n-k}+1=f(k+1)+n+k+3<2n $$ This proves the claim, since then either $|f(k)|<n$ or $|f(k+1)|<n$.