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Proof step 5 from proof abs-peb-empty (previous step)

emptyleb = buf, aehdr ≠ empty → avhdr ≠ empty ∧ 0 = m0 + 2 * EB_PAGE_SIZE ⊦ aehdr = garbage, avhdr = garbage, avhdr.valid?, aehdr.valid?, aehdr = empty ∧ avhdr = empty ∧ buf = mkempbuf(LEB_SIZE)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ emptyleb = mkempbuf(LEB_SIZE)

Rule argument

Additional information

Used simplifier rules: ⊦ empty ≠ garbage; (s)% ⊦ empty ≠ garbage; (s)% ⊦ avhdr ≠ garbage → (avhdr.valid? ↔ avhdr ≠ empty); (ss)% ⊦ aehdr ≠ garbage → (aehdr.valid? ↔ aehdr ≠ empty); (ss)