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Proof step 4 from proof not-valid-peb2-explicit-0 (previous step)

aehdr.valid?, m1 ≠ 0 → m1 ≤ datasize(buf) ∧ m0 = checksum(buf, m1) ⊦ ∃ m3, m4, m2. avidhdr(v, m, n, m1, m0) = avidhdr(v, m, m3, m4, m2) ∧ (m4 ≠ 0 → m4 ≤ datasize(buf) ∧ m2 = checksum(buf, m4))

Heuristic quantifier applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: n, m1, m0

Additional information

Used terms: n, m1, m0