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Proof step 24 from proof reduce-initial (previous step)

pebs[n] = mkpeb(buf0, m0, boolvar0), pebs[n0] = mkpeb(buf1, m1, boolvar1), n0 < # pebs, n < # pebs ∧ # buf0 = PEB_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)))) → # buf = PEB_SIZE ∧ (¬ boolvar → m ≤ PEB_SIZE ∧ is-aligned(m, EB_PAGE_SIZE) ∧ range=(buf, m, emptypeb, m, PEB_SIZE - m)) ∧ (¬ boolvar → m = EB_PAGE_SIZE ∧ is-encoded(echdr(n1), subarray(buf, 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)))) ⊦ n0 = n, # buf1 = PEB_SIZE ∧ (¬ boolvar1 → m1 ≤ PEB_SIZE ∧ is-aligned(m1, EB_PAGE_SIZE) ∧ range=(buf1, m1, emptypeb, m1, PEB_SIZE - m1)) ∧ (¬ boolvar1 → m1 = EB_PAGE_SIZE ∧ (∃ n1. is-encoded(echdr(n1), subarray(buf1, 0, EB_PAGE_SIZE))))

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: n0

Additional information

Used simplifier rules: ⊦ mkpeb(buf, m, boolvar).data = buf; (s)% ⊦ mkpeb(buf, m, boolvar).bad ↔ boolvar; (s)% ⊦ mkpeb(buf, m, boolvar).fill = m; (s)% ⊦ is-encoded(ehdr, buf) → (is-encoded(ehdr0, buf) ↔ ehdr = ehdr0); (s)% ⊦ echdr(m) = echdr(m0) ↔ m = m0; (s) Used terms: n0