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Proof step 13 from proof valid-eraseq-stronger-no-maxsqnum (previous step)

avhdr.sqn = valid-sqns(v, m, apebs).max, apebs[n] = mkpeb(aehdr, avhdr, buf, m0, boolvar), apebs[m3] = mkpeb(aehdr1, avhdr1, buf1, m4, boolvar1), m3 < # apebs, n < # apebs, valid-eraseq-stronger(apebs, q0), valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)), ∀ avhdr0, v0, m1, apebs0, n0, aehdr0, buf0, m2, boolvar0. avhdr0.sqn ≠ valid-sqns(v0, m1, apebs0).max ∨ apebs0[n0] ≠ mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0) ∨ ¬ n0 < # apebs0 ∨ ¬ valid-eraseq-stronger(apebs0, q0) ∨ ¬ valid-peb(v0, m1, mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0)) ∨ ¬ n0 ∈ q0, ∀ v, m. (valid-peb(v, m, mkpeb(aehdr1, avhdr1, buf1, m4, boolvar1)) ↔ lref = v × m) ∧ (lref = v × m → avhdr1.sqn ≠ valid-sqns(v, m, apebs).max), n ∈ q0 ∨ n = m3 ⊦ m3 ∈ q0

User applied rule all left

And obtained the following new goals

Proof step 14:
avhdr.sqn = valid-sqns(v, m, apebs).max, apebs[m3] = mkpeb(aehdr1, avhdr1, buf1, m4, boolvar1), apebs[n] = mkpeb(aehdr, avhdr, buf, m0, boolvar), n < # apebs, m3 < # apebs, valid-eraseq-stronger(apebs, q0), valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)), ∀ v, m. (valid-peb(v, m, mkpeb(aehdr1, avhdr1, buf1, m4, boolvar1)) ↔ lref = v × m) ∧ (lref = v × m → avhdr1.sqn ≠ valid-sqns(v, m, apebs).max), n ∈ q0 ∨ n = m3, avhdr.sqn ≠ valid-sqns(v, m, apebs).max ∨ apebs[n] ≠ mkpeb(aehdr, avhdr, buf, m0, boolvar) ∨ ¬ n < # apebs ∨ ¬ valid-eraseq-stronger(apebs, q0) ∨ ¬ valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)) ∨ ¬ n ∈ q0 ⊦ m3 ∈ q0

Rule argument

pos: left 8 terms: avhdr, v, m, apebs, n, aehdr, buf, m0, boolvar abort: F discard: T computed: T

Additional information

Used terms: avhdr, v, m, apebs, n, aehdr, buf, m0, boolvar