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Proof step 27 from proof valid-eraseq-stronger-next-append-lref (previous step)

valid-peb(v0, m1, mkpeb(aehdr0, avhdr0, buf0, m3, boolvar0)) ↔ lref = v0 × m1, apebs[0 + m2] = mkpeb(aehdr1, avhdr1, buf1, m4, boolvar1), apebs[m2] = mkpeb(aehdr0, avhdr0, buf0, m3, boolvar0), apebs[m] = mkpeb(aehdr, avhdr, buf, m0, boolvar), avhdr.sqn < valid-sqns(v, n, apebs, m).max, m2 < # apebs, m < # apebs, m2 < m, m2 + 1 < # apebs, m2 < m + 1, valid-eraseq-stronger(subarray(apebs, 0, m + 1), q0 + eq-entry(m, v × n) '), valid-eraseq-stronger(subarray(apebs, 0, m), q0), valid-peb(v, n, mkpeb(aehdr, avhdr, buf, m0, boolvar)), valid-peb(v0, m1, mkpeb(aehdr0, avhdr0, buf0, m3, boolvar0)) → avhdr0.sqn ≠ valid-sqns(v0, m1, apebs, m).max ⊦ m2 = m, valid-sqns(v, n, apebs, m) = ∅, m2 ∈ q0, ( valid-peb(v0, m1, mkpeb(aehdr0, avhdr0, buf0, m3, boolvar0)) → avhdr1.sqn ≠ (valid-sqns(v0, m1, apebs, m) ∪ (v = v0 ∧ n = m1 ⊃ {avhdr.sqn} ;∅)).max) ∧ (valid-peb(v0, m1, mkpeb(aehdr1, avhdr1, buf1, m4, boolvar1)) ↔ valid-peb(v0, m1, mkpeb(aehdr0, avhdr0, buf0, m3, boolvar0)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
avhdr0.sqn = (valid-sqns(v0, m1, apebs, m) ∪ (v = v0 ∧ n = m1 ⊃ {avhdr.sqn} ;∅)).max, apebs[m2] = mkpeb(aehdr0, avhdr0, buf0, m3, false), apebs[m] = mkpeb(aehdr, avhdr, buf, m0, boolvar), m2 + 1 < # apebs, m < # apebs, m2 < # apebs, avhdr.sqn < valid-sqns(v, n, apebs, m).max, m2 < m, valid-eraseq-stronger(subarray(apebs, 0, m), q0), valid-eraseq-stronger(subarray(apebs, 0, m + 1), q0 + eq-entry(m, v × n) '), valid-peb(v0, m1, mkpeb(aehdr0, avhdr0, buf0, m3, false)), valid-peb(v, n, mkpeb(aehdr, avhdr, buf, m0, boolvar)) ⊦ avhdr0.sqn = valid-sqns(v0, m1, apebs, m).max, m2 = m, valid-sqns(v, n, apebs, m) = ∅, m2 ∈ q0

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, true)); (s)%⊦ valid-peb(v0, m0, apeb) → (valid-peb(v1, m1, apeb) ↔ v0 = v1 ∧ m0 = m1); (s)%⊦ mkpeb(aehdr, avhdr, buf, m, boolvar) = mkpeb(aehdr0, avhdr0, buf0, m0, boolvar0) ↔ aehdr = aehdr0 ∧ avhdr = avhdr0 ∧ buf = buf0 ∧ m = m0 ∧ (boolvar ↔ boolvar0); (s)% ⊦ m ≠ n → (m < n + 1 ↔ m < n); (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)