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Proof step 14 from proof in-eraseq-strong-not-valid (previous step)

avhdr.sqn = valid-sqns(v, m, apebs).max, apebs[m3] = mkpeb(aehdr1, avhdr1, buf1, m4, boolvar1), apebs[n] = mkpeb(aehdr, avhdr, buf, m0, boolvar), m3 < # apebs, m < # vols[v], valid-eraseq-strong(apebs, to-vtbl(vols), q0), valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)), v ∈ vols, ∀ avhdr0, v0, m1, apebs0, n0, aehdr0, buf0, m2, boolvar0, vols0. avhdr0.sqn ≠ valid-sqns(v0, m1, apebs0).max ∨ apebs0[n0] ≠ mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0) ∨ ¬ m1 < # vols0[v0] ∨ ¬ valid-eraseq-strong(apebs0, to-vtbl(vols0), q0) ∨ ¬ valid-peb(v0, m1, mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0)) ∨ ¬ n0 ∈ q0 ∨ ¬ v0 ∈ vols0, ∀ v0, m1. (valid-peb(v0, m1, mkpeb(aehdr1, avhdr1, buf1, m4, boolvar1)) ↔ lref = v0 × m1) ∧ (lref = v0 × m1 ∧ v0 ∈ vols ∧ m1 < # vols[v0] → avhdr1.sqn ≠ valid-sqns(v0, m1, apebs).max), n = m3 ∨ n ∈ q0 ⊦ m3 ∈ q0

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 11

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