Description: Using the recursion formula
"(n+1)-mintru-(m+1)" <-> if- ( ph , "n-mintru-m" , "n-mintru-(m+1)" )
for "2-mintru-1" (meaning "at least 1 out of 2 inputs is true") by plugging in n = 1, m = 0, and simplifying. The expression "1-mintru-0" is a base case (meaning at least zero inputs out of 1 are true), evaluating to T. , and wl-1mintru1 shows "1-mintru-1" is equivalent to the only input.
Negating an "n-mintru1" operation means: All n inputs ph .. th are false. This is also conveniently expressed as -. ( ph \/ .. \/ th ) , in accordance with the result here. (Contributed by Wolf Lammen, 10-May-2024)
Ref | Expression | ||
---|---|---|---|
Assertion | wl-2mintru1 | |- ( if- ( ps , T. , ch ) <-> ( ps \/ ch ) ) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | dfifp3 | |- ( if- ( ps , T. , ch ) <-> ( ( ps -> T. ) /\ ( ps \/ ch ) ) ) |
|
2 | trud | |- ( ps -> T. ) |
|
3 | 2 | bitru | |- ( ( ps -> T. ) <-> T. ) |
4 | 3 | anbi1i | |- ( ( ( ps -> T. ) /\ ( ps \/ ch ) ) <-> ( T. /\ ( ps \/ ch ) ) ) |
5 | truan | |- ( ( T. /\ ( ps \/ ch ) ) <-> ( ps \/ ch ) ) |
|
6 | 1 4 5 | 3bitri | |- ( if- ( ps , T. , ch ) <-> ( ps \/ ch ) ) |