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Proof step 14 from proof datasize-le-h (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
(buf[m1] = empty ⊃ datasize(buf, m1);m1 + 1) < m, m1 < # buf, n < # buf, (buf[m1] = empty ⊃ datasize(buf, m1);m1 + 1) < # buf, n ≤ m1, m ≤ n + 1, (buf[m1] = empty ⊃ datasize(buf, m1);m1 + 1) ≤ n, m ≤ # buf, (buf[m1] = empty ⊃ datasize(buf, m1);m1 + 1) ≤ 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: le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → n ≠ 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)