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Proof step 21 from proof isempty-append (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
n < # buf0 + # buf1, ∀ n. n < # buf0 → buf0[n] = empty, ∀ n. n < # buf1 → buf1[n] = empty ⊦ (buf0 + buf1)[n] = empty

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic