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Proof step 3 from proof in-eraseq-strong-valid (previous step)

valid-eraseq-strong(apebs, to-vtbl(vols), q), eq-entry(n, v × m) ∈ q ⊦ valid-peb(v, m, apebs[n]) ∧ (v ∈ vols ∧ m < # vols[v] → apebs[n].vidhdr.sqn ≠ valid-sqns(v, m, apebs).max)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
valid-eraseq-strong(apebs, to-vtbl(vols), eqent + q0), eq-entry(n, v × m) ∈ eqent + q0, ∀ apebs, vols, n, v, m. valid-eraseq-strong(apebs, to-vtbl(vols), q0) ∧ eq-entry(n, v × m) ∈ q0 → valid-peb(v, m, apebs[n]) ∧ (v ∈ vols ∧ m < # vols[v] → apebs[n].vidhdr.sqn ≠ valid-sqns(v, m, apebs).max) ⊦ valid-peb(v, m, apebs[n]) ∧ (v ∈ vols ∧ m < # vols[v] → apebs[n].vidhdr.sqn ≠ valid-sqns(v, m, apebs).max)

Rule argument

term: q

Additional information

Used simplifier rules: queue freely generated by [], +;;% ⊦ ¬ eqent ∈ []; (s)