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Proof step 16 from proof datasize-le-h (previous step)

buf[m1] = empty, (buf[m1] = empty ⊃ datasize(buf, m1);m1 + 1) < # buf, n < # buf, m1 < # buf, datasize(buf, m1) < m, (buf[m1] = empty ⊃ datasize(buf, m1);m1 + 1) ≤ m1, m ≤ # buf, (buf[m1] = empty ⊃ datasize(buf, m1);m1 + 1) ≤ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
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

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat