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Proof step 6 from proof not-valid-peb2-explicit-0 (previous step)

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

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 7:
m1 = 0, aehdr.valid? ⊦ ∃ v0, m4, m5, m2, m3. avidhdr(v, m, n, m1, m0) = avidhdr(v0, m4, m5, m2, m3) ∧ (m2 ≠ 0 → m2 ≤ datasize(buf) ∧ m3 = checksum(buf, m2))
Proof step 9:
m0 = checksum(buf, m1), aehdr.valid?, m1 ≤ datasize(buf) ⊦ ∃ v0, m4, m5, m2, m3. avidhdr(v, m, n, m1, m0) = avidhdr(v0, m4, m5, m2, m3) ∧ (m2 ≠ 0 → m2 ≤ datasize(buf) ∧ m3 = checksum(buf, m2))

Rule argument

pos: left 2

Proof step 6 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