Induction, 0'1 and 1's sequence fun question

Question

1. 010 can we generated.
2. If s is a sequence which can be generated by these rules, then 01s, 10s, 0s1, 1s0, s01, and s10 can all be generated.

-Prove (by induction?!) that in any sequence generated by these rules, there are more 0's than 1's

Without using the principle of induction my best guess was that rule 1 is the base case hence there will always be one 0 more than 1. The order in which the sequences are displayed do not matter because they just add one 0 and one 1 to the sequence s. BUT the question demands induction.. I can't see any other way to explain it rather than the way I just did.

PART 2 (if you're bored :p) Again, consider sequences of 0's and 1's, this time generated according to the following rules: 10 can be generated.

01 can be generated.

If s is a sequence which can be generated by these rules, then 0s1 and 1s0 can be generated by these rules.

If s is a sequence which can be generated by these rules, then the sequence ss (s followed by another copy of s) can be generated.

If s is a sequence which can be generated by these rules, and s is of the form t00 or t11 for some smaller sequence t, then t itself can be generated.

Prove that no sequence with an odd length can be generated by these rules.

Answer 1

In any induction proof there are always these elements:

• A sentence (which is to prove) that depends on $n$.
• The base case: the proof that the sentence is valid when $n=1$ (sometimes $0$, sometimes other numbers).
• The inductive step: under the assumption that the sentence is valid for $1\le n\le N$, the proof that is valid also for $n=N+1$.

I think that the point here is identifying these elements in the problem:

• The sentence to prove: every sequence that can be generated (following the rules) in $n$ steps has more zeros than ones.
• Base case: prove that the sentences that can be generated in only one step have more zeros than ones.
• Inductive step: under the assumption that every sequence generated in $N$ steps or less has more zeros than ones, proof that every sequence generated in $N+1$ steps also has.