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Proof step 16 from proof isempty-append (previous step)

∀ n. n < # buf0 + # buf1 → (buf0 + buf1)[n] = empty ⊦ (n < # buf0 → buf0[n] = empty) ∧ (n0 < # buf1 → buf1[n0] = empty)

User applied rule case distinction

And obtained the following new goals

Proof step 17:
n < # buf0, ∀ n. n < # buf0 + # buf1 → (buf0 + buf1)[n] = empty ⊦ buf0[n] = empty
Proof step 19:
n0 < # buf1, ∀ n. n < # buf0 + # buf1 → (buf0 + buf1)[n] = empty ⊦ buf1[n0] = empty

Rule argument

pos: right 1

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