Theorem wl-1mintru2 35660

Description: Using the recursion formula:

"(n+1)-mintru-(m+1)" ↔ if-(𝜑, "n-mintru-m" , "n-mintru-(m+1)" )

for "1-mintru-2" (meaning "at least 2 out of a single input are true") by plugging in n = 0, m = 1, and simplifying. The expressions "0-mintru-1" and "0-mintru-2" are base cases of the recursion, meaning "in a sequence of zero inputs at least 1 / 2 input is true", evaluate both to ⊥.

Since no sequence of inputs has a longer subsequence of whatever property, the resulting ⊥ is to be expected.

Negating a "n-mintru2" operation has an interesting interpretation: 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 = 1) leads to the obvious "In a single input sequence 'at most one is true' always holds". (Contributed by Wolf Lammen, 10-May-2024.)

Assertion

Ref	Expression
wl-1mintru2	⊢ (if-(𝜒, ⊥, ⊥) ↔ ⊥)

Proof of Theorem wl-1mintru2

Step	Hyp	Ref	Expression
1		ifpid 1075	1 ⊢ (if-(𝜒, ⊥, ⊥) ↔ ⊥)

Syntax hints:   ↔ wb 205  if-wif 1060  ⊥wfal 1551

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 206  df-an 397  df-or 845  df-ifp 1061

This theorem is referenced by: (None)