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Proof step 1 from proof not-valid-peb2-explicit-0 (conclusion)

⊦ valid-peb(mkpeb(aehdr, avidhdr(v, m, n, n0, n2), buf, n1, boolvar)) ↔ valid-peb(v, m, mkpeb(aehdr, avidhdr(v, m, n, n0, n2), buf, n1, boolvar))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
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)))

Rule argument

Additional information

Used simplifier rules: valid-peb0: ⊦ valid-peb(apeb) ↔ (∃ v, m. valid-peb(v, m, apeb));%valid-peb2: ⊦ valid-peb(v, m, apeb) ↔ ¬ apeb.bad ∧ apeb.echdr.valid? ∧ (∃ m0, m1, m2. apeb.vidhdr = avidhdr(v, m, m0, m1, m2) ∧ ( m1 ≠ 0 → m1 ≤ datasize(apeb.data) ∧ m2 = checksum(apeb.data, m1)));% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).data = buf; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).echdr = aehdr; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).bad ↔ boolvar; (s)