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Proof step 2 from proof not-valid-peb2-explicit-0 (previous step)

aehdr.valid? ⊦ (∃ v0, m0, m1, m2, m3. avidhdr(v, m, n, n0, n2) = avidhdr(v0, m0, m1, m2, m3) ∧ (m2 ≠ 0 → m2 ≤ datasize(buf) ∧ m3 = checksum(buf, m2))) ↔ (∃ m0, m1, m2. avidhdr(v, m, n, n0, n2) = avidhdr(v, m, m0, m1, m2) ∧ (m1 ≠ 0 → m1 ≤ datasize(buf) ∧ m2 = checksum(buf, m1)))

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 3:
aehdr.valid?, ∃ v0, m0, m1, m2, m3. avidhdr(v, m, n, n0, n2) = avidhdr(v0, m0, m1, m2, m3) ∧ (m2 ≠ 0 → m2 ≤ datasize(buf) ∧ m3 = checksum(buf, m2)) ⊦ ∃ m0, m1, m2. avidhdr(v, m, n, n0, n2) = avidhdr(v, m, m0, m1, m2) ∧ (m1 ≠ 0 → m1 ≤ datasize(buf) ∧ m2 = checksum(buf, m1))
Proof step 5:
aehdr.valid?, ∃ m0, m1, m2. avidhdr(v, m, n, n0, n2) = avidhdr(v, m, m0, m1, m2) ∧ (m1 ≠ 0 → m1 ≤ datasize(buf) ∧ m2 = checksum(buf, m1)) ⊦ ∃ v0, m0, m1, m2, m3. avidhdr(v, m, n, n0, n2) = avidhdr(v0, m0, m1, m2, m3) ∧ (m2 ≠ 0 → m2 ≤ datasize(buf) ∧ m3 = checksum(buf, m2))

Rule argument

pos: right 1

Proof step 2 from proof not-valid-peb2-explicit-0 (previous step)

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Rule argument

Additional information