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Proof step 7 from proof datasize-at-h (previous step)

n < # buf, m0 < # buf, (buf[m0] = empty ⊃ datasize(buf, m0);m0 + 1) < # buf, (buf[m0] = empty ⊃ datasize(buf, m0);m0 + 1) ≤ n, n ≤ m0, (buf[m0] = empty ⊃ datasize(buf, m0);m0 + 1) ≤ m0, ∀ buf, n. datasize(buf, m0) < # buf ∧ n < m0 ∧ datasize(buf, m0) < m0 ∧ n < # buf ∧ m0 ≤ # buf ∧ datasize(buf, m0) ≤ n → buf[n] = empty ⊦ buf[n] = empty

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 8:
buf[m0] = empty, (buf[m0] = empty ⊃ datasize(buf, m0);m0 + 1) < # buf, m0 < # buf, n < # buf, (buf[m0] = empty ⊃ datasize(buf, m0);m0 + 1) ≤ m0, n ≤ m0, datasize(buf, m0) ≤ n, ∀ buf, n. datasize(buf, m0) < # buf ∧ n < m0 ∧ datasize(buf, m0) < m0 ∧ n < # buf ∧ m0 ≤ # buf ∧ datasize(buf, m0) ≤ n → buf[n] = empty ⊦ buf[n] = empty

Rule argument

pos: left 4

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