Theorem wl-2mintru2 37982

Description: Using the recursion formula

"(n+1)-mintru-(m+1)" ↔ if-(𝜑, "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 37979 and wl-1mintru2 37980 to see that "1-mintru-1" / "1-mintru-2" evaluate to 𝜒 / ⊥ 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 (𝜑 ⊼ 𝜓), 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.)

Assertion

Ref	Expression
wl-2mintru2	⊢ (if-(𝜓, 𝜒, ⊥) ↔ (𝜓 ∧ 𝜒))

Proof of Theorem wl-2mintru2

Step	Hyp	Ref	Expression
1		dfifp7 1081	. 2 ⊢ (if-(𝜓, 𝜒, ⊥) ↔ ((⊥ → 𝜓) → (𝜓 ∧ 𝜒)))
2		falim 1577	. . 3 ⊢ (⊥ → 𝜓)
3	2	a1bi 364	. 2 ⊢ ((𝜓 ∧ 𝜒) ↔ ((⊥ → 𝜓) → (𝜓 ∧ 𝜒)))
4	1, 3	bitr4i 280	1 ⊢ (if-(𝜓, 𝜒, ⊥) ↔ (𝜓 ∧ 𝜒))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 399  if-wif 1074  ⊥wfal 1572

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-ifp 1075  df-tru 1563  df-fal 1573

This theorem is referenced by: (None)