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Proof step 4 from proof abs-peb-empty (previous step)

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

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information