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Proof step 25 from proof datasize-further (previous step)

(buf[n] = empty ⊃ datasize(buf, n);n + 1) = datasize(buf, n1), n < n1, n < # buf, ∀ n1, buf. n ≤ n1 ∧ n ≤ # buf → (datasize(buf, n) = datasize(buf, n1) ↔ (∀ n2. n ≤ n2 ∧ n2 < n1 → buf[n2] = empty)) ⊦ ∀ n0. n < n0 ∧ n0 < n1 → buf[n0] = empty

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
(buf[n] = empty ⊃ datasize(buf, n);n + 1) = datasize(buf, n1), n < n1, n < # buf, n0 < n1, n < n0, n + 1 < n1, ∀ n1, buf. n ≤ n1 ∧ n ≤ # buf → (datasize(buf, n) = datasize(buf, n1) ↔ (∀ n2. n ≤ n2 ∧ n2 < n1 → buf[n2] = empty)) ⊦ buf[n0] = empty

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic