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Proof step 5 from proof valid-peb-imp-valid-peb0 (previous step)

aehdr.valid?, m1 ≠ 0 → m1 ≤ datasize(buf) ∧ m2 = checksum(buf, m1) ⊦ ∃ v0, m3, m4, m5, m6. avidhdr(v, m, m0, m1, m2) = avidhdr(v0, m3, m4, m5, m6) ∧ (m5 ≠ 0 → m5 ≤ datasize(buf) ∧ m6 = checksum(buf, m5))

Heuristic quantifier applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: v, m, m0, m1, m2

Additional information

Used terms: v, m, m0, m1, m2