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Proof step 8 from proof isempty-empbuf (previous step)

⊦ (∀ n. n < # buf → buf[n] = empty) ↔ # buf = # mkempbuf(# buf) ∧ (∀ n. n < # buf → buf[n] = mkempbuf(# buf)[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
n < # buf, ∀ n. n < # buf → buf[n] = empty ⊦ buf[n] = empty

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → mkempbuf(n)[m] = empty; (s)% ⊦ # mkempbuf(n) = n; (s)