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Proof step 2 from proof downcase-B (
previous step
)
⊦ pos('A', sortedchars) ≤ pos(char, sortedchars) ∧ pos(char, sortedchars) ≤ pos('Z', sortedchars) → downcase(char) = s→l(sortedchars)[pos(char, sortedchars) + (pos('a', sortedchars) - pos('A', sortedchars))]
Lemma
Char-downcase-yes