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Proof step 25 from proof isempty-append (previous step)

n < # buf0 + # buf1, ∀ n. n < # buf0 → buf0[n] = empty, ∀ n. n < # buf1 → buf1[n] = empty ⊦ (n < # buf0 ⊃ buf0[n];buf1[n - # buf0]) = empty

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 26:
n < # buf0 + # buf1, n < # buf0, ∀ n. n < # buf1 → buf1[n] = empty, ∀ n. n < # buf0 → buf0[n] = empty ⊦ buf0[n] = empty
Proof step 28:
n < # buf0 + # buf1, ∀ n. n < # buf1 → buf1[n] = empty, ∀ n. n < # buf0 → buf0[n] = empty ⊦ buf1[n - # buf0] = empty, n < # buf0

Rule argument

pos: right 1

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