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Proof step 11 from proof upcase-a (previous step)

58 ≤ pos('z', sortedchars) ⊦ s→l(sortedchars)[58 - (58 - pos('A', sortedchars))] = 'A'

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
58 ≤ pos('z', sortedchars) ⊦ s→l(sortedchars)[58 - (58 - pos('A', sortedchars))] = 'A', pos('z', sortedchars) = 0

Rule argument

Additional information

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