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Proof step 12 from proof datasize-le-h (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
m1 < # buf, m ≤ m1 + 1, m ≤ # buf, ∀ buf, m. m1 ≤ # buf → (m ≤ datasize(buf, m1) ↔ (∃ n. m ≤ n + 1 ∧ n < m1 ∧ buf[n] ≠ empty)) ∨ m = 0 ⊦ buf[m1] = empty, m = 0, ∃ n. m ≤ n + 1 ∧ ¬ m1 < n ∧ 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%le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat