Take the machine for BB(n) with non-end states {1,2,3,…n} and end state E.
In its program, change every transition that goes to state E to go to a new state n+1 instead.
Add transitions “when in state n+1 and on a 1, move right” and “when in state n+1 and on a 0, write a 1 and stop”.
Doesn’t that give you a Turing machine with n+1 states that stops after writing one more 1 than BB(n) writes, thus proving that BB(n+1) is at least one more than BB(n+1)?
In its program, change every transition that goes to state E to go to a new state n+1 instead.
Add transitions “when in state n+1 and on a 1, move right” and “when in state n+1 and on a 0, write a 1 and stop”.
Doesn’t that give you a Turing machine with n+1 states that stops after writing one more 1 than BB(n) writes, thus proving that BB(n+1) is at least one more than BB(n+1)?