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Proof step 11 from proof downcase-A (previous step)

28 ≤ pos('Z', sortedchars) ⊦ s→l(sortedchars)[28 + (pos('a', sortedchars) - 28)] = 'a'

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
28 ≤ pos('Z', sortedchars) ⊦ s→l(sortedchars)[28 + (pos('a', sortedchars) - 28)] = 'a', pos('Z', sortedchars) = 0

Rule argument

Additional information

Used simplifier rules: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)