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Proof step 17 from proof valid-eraseq-strong-notinvols (previous step)

eqent = eq-entry(m0, lref), m0 < # apebs, n < # apebs, valid-eraseq-strong(apebs, vtbl, q0), valid-peb(v, m, apebs[n]), ∀ v, m. (valid-peb(v, m, apebs[m0]) ↔ lref = v × m) ∧ (lref = v × m ∧ v ∈ vtbl ∧ m < vtbl[v] → apebs[m0].vidhdr.sqn ≠ valid-sqns(v, m, apebs).max), ∀ n, apebs, vtbl, v, m. n < # apebs ∧ valid-eraseq-strong(apebs, vtbl, q0) ∧ valid-peb(v, m, apebs[n]) ∧ (¬ v ∈ vtbl ∨ ¬ m < vtbl[v]) → valid-eraseq-strong(apebs, vtbl, q0 + eq-entry(n, v × m) ') ∨ n ∈ q0, ¬ v ∈ vtbl ∨ ¬ m < vtbl[v] ⊦ n ∈ q0, n ∈ eqent ', m0 ∈ q0, ¬ m0 ∈ q0 + eq-entry(n, v × m) ' ∧ valid-eraseq-strong(apebs, vtbl, q0 + eq-entry(n, v × m) ')

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
n < # apebs, m0 < # apebs, valid-eraseq-strong(apebs, vtbl, q0), valid-peb(v, m, apebs[n]), ∀ n, apebs, vtbl, v, m. n < # apebs ∧ valid-eraseq-strong(apebs, vtbl, q0) ∧ valid-peb(v, m, apebs[n]) ∧ (¬ v ∈ vtbl ∨ ¬ m < vtbl[v]) → valid-eraseq-strong(apebs, vtbl, q0 + eq-entry(n, v × m) ') ∨ n ∈ q0, ∀ v, m. (valid-peb(v, m, apebs[m0]) ↔ lref = v × m) ∧ (lref = v × m ∧ v ∈ vtbl ∧ m < vtbl[v] → apebs[m0].vidhdr.sqn ≠ valid-sqns(v, m, apebs).max), ¬ v ∈ vtbl ∨ ¬ m < vtbl[v] ⊦ n = m0, valid-eraseq-strong(apebs, vtbl, q0 + eq-entry(n, v × m) '), n ∈ q0, m0 ∈ q0

Rule argument

Additional information

Used simplifier rules: a: ⊦ (q + q0) + q1 = q + q0 + q1; from specification list[erasequeue-base]% ⊦ n ≠ n0 → (n ∈ q + eq-entry(n0, lref) ' ↔ n ∈ q); (s)% ⊦ n ∈ eq-entry(m, lref) ' ↔ n = m; (s)