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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

User applied rule structural induction

And obtained the following new goals

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

Rule argument

indvar: m

Additional information

Used simplifier rules: Fill-zero: ⊦ fill(ar, a, m, 0) = ar;%nat freely generated by 0, +1;;% ⊦ ¬ m < 0; (s)