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Proof step 14 from proof downcase-A (previous step)

28 ≤ 53 ⊦ s→l(sortedchars)[28 + (pos('a', sortedchars) - 28)] = 'a', 53 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
⊦ s→l(sortedchars)[28 + (pos('a', sortedchars) - 28)] = 'a'

Rule argument

Additional information