Description: Using the recursion formula:
"(n+1)-mintru-(m+1)" <-> if- ( ph , "n-mintru-m" , "n-mintru-(m+1)" )
for "1-mintru-1" (meaning "at least 1 out of 1 input is true") by plugging in n = 0, m = 0, and simplifying. The expressions "0-mintru-0" and "0-mintru-1" are base cases of the recursion, meaning "in a sequence of zero inputs, at least 0 / 1 input is true", respectively equvalent to T. / F. .
Negating an "n-mintru1" operation means: All n inputs ph .. th are false. This is also conveniently expressed as -. ( ph \/ .. \/ th ) . Applying this idea here (n = 1) yields the obvious result that in an input sequence of size 1 only then all will be false, if its single input is. (Contributed by Wolf Lammen, 10-May-2024)
Ref | Expression | ||
---|---|---|---|
Assertion | wl-1mintru1 | |- ( if- ( ch , T. , F. ) <-> ch ) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | tbtru | |- ( ch <-> ( ch <-> T. ) ) |
|
2 | 1 | biimpi | |- ( ch -> ( ch <-> T. ) ) |
3 | nbfal | |- ( -. ch <-> ( ch <-> F. ) ) |
|
4 | 3 | biimpi | |- ( -. ch -> ( ch <-> F. ) ) |
5 | 2 4 | casesifp | |- ( ch <-> if- ( ch , T. , F. ) ) |
6 | 5 | bicomi | |- ( if- ( ch , T. , F. ) <-> ch ) |