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Proof step 1 from proof rev-at-h (conclusion)

⊦ m < n ∧ n ≤ # ar → rev(ar, n)[m] = ar[# ar - m + 1]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
m < # ar, n ≤ # ar, m < n ⊦ n = 0, rev(ar, n)[m] = ar[# ar - m + 1], # ar = 0

Rule argument

Additional information

Used simplifier rules: fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1