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Proof step 3 from proof datasize-zero-h (previous step)

∀ buf. datasize(buf, m0) = 0 ↔ (∀ n. n < m0 → buf[n] = empty) ⊦ datasize(buf, m0 +1) = 0 ↔ (∀ n. n < m0 +1 → buf[n] = empty)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1)
Proof step 5:
∀ buf. datasize(buf, m0) = 0 ↔ (∀ n. n < m0 → buf[n] = empty) ⊦ (buf[m0] = empty ⊃ datasize(buf, m0);m0 + 1) = 0 ↔ (∀ n. ¬ m0 < n → buf[n] = 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)