Metamath Proof Explorer


Theorem wl-2mintru2

Description: Using the recursion formula

"(n+1)-mintru-(m+1)" <-> if- ( ph , "n-mintru-m" , "n-mintru-(m+1)" )

for "2-mintru-2" (meaning "2 out of 2 inputs are true") by plugging in n = 1, m = 1, and simplifying. See wl-1mintru1 and wl-1mintru2 to see that "1-mintru-1" / "1-mintru-2" evaluate to ch / F. respectively.

Negating a "n-mintru2" operation means 'at most one input is true', so all inputs exclude each other mutually. Such an exclusion is expressed by a NAND operation ( ph -/\ ps ) , not by a XOR. Applying this idea here (n = 2) yields the expected NAND in case of a pair of inputs. (Contributed by Wolf Lammen, 10-May-2024)

Ref Expression
Assertion wl-2mintru2 ( if- ( 𝜓 , 𝜒 , ⊥ ) ↔ ( 𝜓𝜒 ) )

Proof

Step Hyp Ref Expression
1 dfifp7 ( if- ( 𝜓 , 𝜒 , ⊥ ) ↔ ( ( ⊥ → 𝜓 ) → ( 𝜓𝜒 ) ) )
2 falim ( ⊥ → 𝜓 )
3 2 a1bi ( ( 𝜓𝜒 ) ↔ ( ( ⊥ → 𝜓 ) → ( 𝜓𝜒 ) ) )
4 1 3 bitr4i ( if- ( 𝜓 , 𝜒 , ⊥ ) ↔ ( 𝜓𝜒 ) )