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Proof step 4 from proof datasize-further (previous step)

0 ≤ # buf, 0 ≤ n1 ⊦ datasize(buf, 0) = datasize(buf, n1) ↔ (∀ n. 0 ≤ n ∧ n < n1 → buf[n] = empty)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ datasize(buf, 0) = 0
Proof step 6:
⊦ 0 = datasize(buf, n1) ↔ (∀ n. n < n1 → buf[n] = empty)

Rule argument

Additional information

Used simplifier rules: datasize-base: ⊦ datasize(buf, 0) = 0;% ⊦ 0 ≤ m; (s)