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Proof step 10 from proof datasize-le-h (previous step)

buf[m1] = empty, m1 < # buf, m ≤ datasize(buf, 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 ⊦ m = 0, ∃ n. m ≤ n + 1 ∧ ¬ m1 < n ∧ buf[n] ≠ empty

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
buf[m1] = empty, m < # buf, n + 1 < # buf, m1 < # buf, n < m1, n < # buf, m ≤ datasize(buf, m1), m ≤ n + 1, m ≤ m1, ∀ buf, m. m1 ≤ # buf → (m ≤ datasize(buf, m1) ↔ (∃ n. m ≤ n + 1 ∧ n < m1 ∧ buf[n] ≠ empty)) ∨ m = 0 ⊦ buf[n] = empty, m = 0, ∃ n. m ≤ n + 1 ∧ ¬ m1 < n ∧ buf[n] ≠ empty

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%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% ⊦ m < n → m ≤ n; (s)