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Proof step 20 from proof datasize-le-h (previous step)

buf[m1] = empty, datasize(buf, m1) < m, m1 < # buf, n < # buf, datasize(buf, m1) < # buf, n ≤ m1, m ≤ n + 1, datasize(buf, m1) ≤ n, m ≤ # buf, datasize(buf, m1) ≤ m1, m1 ≤ # buf → (m ≤ datasize(buf, m1) ↔ (∃ n. m ≤ n + 1 ∧ n < m1 ∧ buf[n] ≠ empty)) ∨ m = 0, ∀ buf, m. m1 ≤ # buf → (m ≤ datasize(buf, m1) ↔ (∃ n. m ≤ n + 1 ∧ n < m1 ∧ buf[n] ≠ empty)) ∨ m = 0 ⊦ buf[n] = empty

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
buf[m1] = empty, datasize(buf, m1) < # buf, n < # buf, m1 < # buf, datasize(buf, m1) < m, datasize(buf, m1) ≤ m1, m ≤ # buf, datasize(buf, m1) ≤ n, m ≤ n + 1, n ≤ m1, ∀ buf, m. m1 ≤ # buf → (m ≤ datasize(buf, m1) ↔ (∃ n. m ≤ n + 1 ∧ n < m1 ∧ buf[n] ≠ empty)) ∨ m = 0 ⊦ buf[n] = empty, ∃ n. m ≤ n + 1 ∧ n < m1 ∧ buf[n] ≠ empty

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)% ⊦ m < n → m ≤ n; (s)