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Proof step 2 from proof at-fill-h0 (previous step)

n < m, n < # ar ⊦ fill(ar, a, 0, m)[n] = a, m = 0, # ar = 0

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ n, ar, a. n < m0 ∧ n < # ar → fill(ar, a, 0, m0)[n] = a ∨ m0 = 0 ∨ # ar = 0, n < m0 +1, n < # ar ⊦ fill(ar, a, 0, m0 +1)[n] = a, m0 +1 = 0, # ar = 0

Rule argument

indvar: m

Additional information

Used simplifier rules: nat freely generated by 0, +1;;