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Proof step 15 from proof datasize-le-h (previous step)

(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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 16:
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

Rule argument

pos: left 1

Additional information

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