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Proof step 1 from proof ext-rec (conclusion)

⊦ ar1 = ar0[n, a] ↔ # ar0 = # ar1 ∧ (∀ m. m < # ar0 → ar1[m] = (n = m ⊃ a;ar0[m]))

User applied rule case distinction

And obtained the following new goals

Proof step 2:
ar1 = ar0[n, a] ⊦ # ar0 = # ar1 ∧ (∀ m. m < # ar0 → ar1[m] = (n = m ⊃ a;ar0[m]))
Proof step 5:
# ar0 = # ar1 ∧ (∀ m. m < # ar0 → ar1[m] = (n = m ⊃ a;ar0[m])) ⊦ ar1 = ar0[n, a]

Rule argument

pos: right 1

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