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Proof step 22 from proof reduce-initial (previous step)

# buf0 = PEB_SIZE, pebs[n] = mkpeb(buf0, m0, boolvar0), n < # pebs, EB_PAGE_SIZE ≤ PEB_SIZE, ¬ boolvar0 → m0 = EB_PAGE_SIZE ∧ is-encoded(echdr(n0), subarray(buf0, 0, EB_PAGE_SIZE)), ¬ boolvar0 → m0 ≤ PEB_SIZE ∧ is-aligned(m0, EB_PAGE_SIZE) ∧ range=(buf0, m0, emptypeb, m0, PEB_SIZE - m0), (¬ boolvar0 → m0 = EB_PAGE_SIZE ∧ (∃ n0. is-encoded(echdr(n0), subarray(buf0, 0, EB_PAGE_SIZE)))) → # buf1 = PEB_SIZE ∧ range=(buf1, EB_PAGE_SIZE, emptypeb, EB_PAGE_SIZE, LEB_SIZE + EB_PAGE_SIZE) ∧ m1 = EB_PAGE_SIZE ∧ is-encoded(echdr(n1), subarray(buf1, 0, EB_PAGE_SIZE)), ∀ n0. n0 < # pebs → # pebs[n0].data = PEB_SIZE ∧ ( ¬ pebs[n0].bad → pebs[n0].fill ≤ PEB_SIZE ∧ is-aligned(pebs[n0].fill, EB_PAGE_SIZE) ∧ range=(pebs[n0].data, pebs[n0].fill, emptypeb, pebs[n0].fill, PEB_SIZE - pebs[n0].fill)) ∧ (¬ pebs[n0].bad → pebs[n0].fill = EB_PAGE_SIZE ∧ (∃ n. is-encoded(echdr(n), subarray(pebs[n0].data, 0, EB_PAGE_SIZE)))) ⊦ ∃ n1. is-encoded(echdr(n1), subarray(buf1, 0, EB_PAGE_SIZE))

Heuristic quantifier applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: n1

Additional information

Used simplifier rules: ⊦ is-encoded(ehdr, buf) → (is-encoded(ehdr0, buf) ↔ ehdr = ehdr0); (s)% ⊦ echdr(m) = echdr(m0) ↔ m = m0; (s) Used terms: n1