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Proof step 28 from proof datasize-append-empty-h (previous step)

m0 + 1 = n0 + # buf0, buf1[m0 - # buf0] = empty, datasize(buf1, n0) = 0, m0 < # buf0 + # buf1, # buf0 ≤ m0, # buf0 ≤ m0 + 1, n0 ≤ # buf1 ⊦ # buf1 = 0, datasize(buf0 + buf1, m0) = datasize(buf0, # buf0), datasize(buf1, m0 - # buf0) = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 29:
⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1)

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [m0, # buf0, n]

Additional information

Used simplifier rules: datasize-rec: ⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1);% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ m + n; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m < n → n ≠ 0; (s)%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m0, # buf0, n