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Proof step 12 from proof datasize-further (previous step)

0 = (buf[n] = empty ⊃ datasize(buf, n);n + 1), n0 ≤ n, ∀ buf. 0 = datasize(buf, n) ↔ (∀ n0. n0 < n → buf[n0] = empty) ⊦ buf[n0] = empty

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 13:
0 = datasize(buf, n), buf[n] = empty, n0 ≤ n, ∀ buf. 0 = datasize(buf, n) ↔ (∀ n0. n0 < n → buf[n0] = empty) ⊦ buf[n0] = empty

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic