grenv: Since you are being pedantic I will have to disagree with your pedantry.

You are welcome....


If you've taken 3 men off, then you have taken a man off. No need to add words like "at least"

Well English is not my native language but i thought "a" = one". That is "here is a pencil" = "here is one pencil". Even if we agree that "a"="one" the problem is not solved. The problem occurs to the different ways we see the "if a man is born off then...."

You are taking the statement:
"If you take a checker you will not lose a gammon" as:
"If you take at least a checker you will not lose a gammon"
while in all mathematical papers and general papers is:
"If you take exactly a checker you will not lose a gammon"

So by "a" what we mean? "Exactly" or "at least"?
Meaning "at least" would make no sense since a single "a" can't mean something more than the number of the object it describes. So why it should mean "exactly"? It doesn't! But in all papers i have read so far the word exactly is used silently to describe that "a man" means "exactly one man".
You may not agree with that but it doesn't matter anyway since it is a matter of different interpretation of statements so it's a subjective matter....

"and you haven't borne any off" is not needed since you would have passed the first test and not gone to the else

Correct! That's why i inserted To show that the word "correct" is not accurate to describe what i did. It was not a precise correction but just a better way to make things more readable.

You are correct. I will say the reason:
Suppose we have different assumtions A,B,C, with B = (D∩A') with D another assumption and C = (A'∩B') and with A∪B∪C = Ω, that from each one we assume X1,X2,X3 respectively, with the following easy way that i tried with my correction to show:
A=>X1 , B=>X2 , C=>X3
So if we have a f that belongs to A then:
-if f belongs to A we have that X1 is correct.
-if f belongs to B we have that X2 is correct.
-if f belongs to C we have that X3 is correct.
so we list this as:
1)A then X1 occurs.
2)B then X2 occurs.
3)C then X3 occurs.

Playbunny used the other equivalent way:
-if f belongs to A we have that X1 is correct.
-If f doesn't belong to A and some other things independed from A occur, we have that X2 is correct.
-if f doesn't belong to A or B we have that X3 is correct.


and we list that as:
1)A then X1 occurs, else
2)D then X2 occurs, else
3)A'∩B' then X3 occurs.

It is equivalent since if 1), that means A, is not correct then it is correct the A'.
So since D is valid then D∩A' is valid too. So X2 would be valid also.
If both of the A or B are not valid, so the A'∩B'; would be valid, then the X3 would be valid also....

So the equivalence of the 2 ways is obvious. The "problem" with the second is that we don't have immediatelly the several discrete cases, although the procedure of the second way is good for programming as we gain some time by not re-checking cases....