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Proof step 6 from proof valid-eraseq-stronger-next (previous step)

eqent = eq-entry(m1, lref), m1 + 1 < # apebs, m < # apebs, m1 < m, m1 < # apebs, valid-eraseq-stronger(subarray(apebs, 0, m), q0), ∀ m, apebs. m < # apebs ∧ valid-eraseq-stronger(subarray(apebs, 0, m), q0) → valid-eraseq-stronger(subarray(apebs, 0, m + 1), q0), ∀ v, m0. (valid-peb(v, m0, apebs[m1]) ↔ lref = v × m0) ∧ (lref = v × m0 → apebs[m1].vidhdr.sqn ≠ valid-sqns(v, m0, apebs, m).max) ⊦ m1 ∈ q0, (valid-peb(v, m0, subarray(apebs, 0, m + 1)[m1]) ↔ lref = v × m0) ∧ ( lref = v × m0 → subarray(apebs, 0, m + 1)[m1].vidhdr.sqn ≠ (valid-sqns(v, m0, apebs, m) ∪ (valid-peb(v, m0, apebs[m]) ⊃ {apebs[m].vidhdr.sqn} ;∅)).max) ∧ valid-eraseq-stronger(subarray(apebs, 0, m + 1), q0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
m1 < # apebs, m < # apebs, m1 < m, m1 + 1 < # apebs, valid-eraseq-stronger(subarray(apebs, 0, m), q0), ∀ m, apebs. m < # apebs ∧ valid-eraseq-stronger(subarray(apebs, 0, m), q0) → valid-eraseq-stronger(subarray(apebs, 0, m + 1), q0), ∀ v, m0. (valid-peb(v, m0, apebs[m1]) ↔ lref = v × m0) ∧ (lref = v × m0 → apebs[m1].vidhdr.sqn ≠ valid-sqns(v, m0, apebs, m).max) ⊦ m1 ∈ q0, (valid-peb(v, m0, subarray(apebs, 0, m + 1)[m1]) ↔ lref = v × m0) ∧ ( lref = v × m0 → subarray(apebs, 0, m + 1)[m1].vidhdr.sqn ≠ (valid-sqns(v, m0, apebs, m) ∪ (valid-peb(v, m0, apebs[m]) ⊃ {apebs[m].vidhdr.sqn} ;∅)).max) ∧ valid-eraseq-stronger(subarray(apebs, 0, m + 1), q0)

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)%transitivity: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification oelem[nat-set]%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic