**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.) |