Metamath Proof Explorer


Theorem wl-2mintru1

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 ) )

Proof

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 ) )