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Proof step 8 from proof not-valid-peb2-explicit-0 (previous step)

aehdr.valid? ⊦ ∃ v0, m4, m5, m2, m3. avidhdr(v, m, n, 0, m0) = avidhdr(v0, m4, m5, m2, m3) ∧ (m2 ≠ 0 → m2 ≤ datasize(buf) ∧ m3 = checksum(buf, m2))

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: v, m, n, 0, m0

Additional information

Used terms: v, m, n, 0, m0