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Proof step 3 from proof abs-fillcount-empty-echdr (previous step)

# pebs = # apebs, apebs[n] = mkpeb(empty, avhdr, buf, m, false), pebs[n] = mkpeb(buf0, m0, false), n < # apebs, ∀ n0. n0 < # apebs → (pebs[n0].bad ↔ apebs[n0].bad) ∧ ( ¬ pebs[n0].bad → pebs[n0].fill = (apebs[n0].echdr = empty ⊃ 0;(apebs[n0].vidhdr = empty ⊃ EB_PAGE_SIZE;apebs[n0].fill + 2 * EB_PAGE_SIZE)) ∧ (apebs[n0].echdr = empty → subarray(pebs[n0].data, 0, EB_PAGE_SIZE) = emptypage) ∧ (apebs[n0].echdr.valid? → is-encoded(echdr(apebs[n0].echdr.ec), subarray(pebs[n0].data, 0, EB_PAGE_SIZE))) ∧ (apebs[n0].echdr = garbage → is-garbage-ec(subarray(pebs[n0].data, 0, EB_PAGE_SIZE)) ∧ subarray(pebs[n0].data, 0, EB_PAGE_SIZE) ≠ emptypage) ∧ (apebs[n0].vidhdr = empty → subarray(pebs[n0].data, EB_PAGE_SIZE, EB_PAGE_SIZE) = emptypage) ∧ ( apebs[n0].vidhdr.valid? → is-encoded (vidhdr(apebs[n0].vidhdr.vol, apebs[n0].vidhdr.leb, apebs[n0].vidhdr.sqn, apebs[n0].vidhdr.size, apebs[n0].vidhdr.checksum), subarray(pebs[n0].data, EB_PAGE_SIZE, EB_PAGE_SIZE))) ∧ ( apebs[n0].vidhdr = garbage → is-garbage-vid(subarray(pebs[n0].data, EB_PAGE_SIZE, EB_PAGE_SIZE)) ∧ subarray(pebs[n0].data, EB_PAGE_SIZE, EB_PAGE_SIZE) ≠ emptypage) ∧ subarray(pebs[n0].data, 2 * EB_PAGE_SIZE, LEB_SIZE) = apebs[n0].data) ⊦ m0 = 0

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: n

Additional information

Used simplifier rules: ⊦ mkpeb(buf, m, boolvar).bad ↔ boolvar; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).bad ↔ boolvar; (s)% ⊦ mkpeb(buf, m, boolvar).fill = m; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).echdr = aehdr; (s) Used terms: n