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Proof step 8 from proof datasize-further (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1)
Proof step 10:
∀ buf. 0 = datasize(buf, n) ↔ (∀ n0. n0 < n → buf[n0] = empty) ⊦ 0 = (buf[n] = empty ⊃ datasize(buf, n);n + 1) ↔ (∀ n0. ¬ n < n0 → buf[n0] = empty)

Rule argument

Additional information

Used simplifier rules: datasize-rec: ⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1);% ⊦ m < n + 1 ↔ ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)