Metamath Proof Explorer

Theorem wl-1mintru2

Description: Using the recursion formula:

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

Since no sequence of inputs has a longer subsequence of whatever property, the resulting F. 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 ( ph -/\ ps ) , 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)

		Ref	Expression
	Assertion	wl-1mintru2	$⊢ if- (χ, ⊥, ⊥) \leftrightarrow ⊥$

Proof

Step	Hyp	Ref	Expression
1		ifpid	$⊢ if- (χ, ⊥, ⊥) \leftrightarrow ⊥$