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Proof step 62 from proof valid-peb-write-valid-peb (previous step)

# buf = LEB_SIZE, LEB_SIZE = n1 + m0, LEB_SIZE = n0 + m, apebs[n] = mkpeb(aehdr, avidhdr(v, m2, m4, m3, checksum(buf, m3)), buf, m, false), aehdr.valid?, n4 < n1, m0 < LEB_SIZE, n < # apebs, m < LEB_SIZE, is-aligned(m, EB_PAGE_SIZE), m ≤ n4 + m0, m ≤ m0, m3 ≤ datasize(buf), m3 ≤ LEB_SIZE, n0 ≤ (m0 - m) + n4 ⊦ buf[n4 + m0] = empty, aehdr = empty, m3 = 0, n0 = 0, m3 ≤ datasize(subarray(buf, 0, m0) + buf1) ∧ checksum(buf, m3) = checksum(subarray(buf, 0, m0) + buf1, m3)

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [m0, m, n2]

Additional information

Used simplifier rules: ⊦ m ≤ m + n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%eqs: ⊦ m = n + n0 ∧ m = n + n1 → n0 = n1; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m < n → n ≠ 0; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m0, m, n2