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Proof step 6 from proof valid-global-sqn-invalidate (previous step)

m < # apebs, n ≤ (valid-sqns(v, m0, apebs) \ (valid-peb(v, m0, apebs[m]) ⊃ {apebs[m].vidhdr.sqn} ;∅)).max, ∀ v, m0. valid-sqns(v, m0, apebs) ≠ ∅ → valid-sqns(v, m0, apebs).max < n ⊦ valid-sqns(v, m0, apebs) \ (valid-peb(v, m0, apebs[m]) ⊃ {apebs[m].vidhdr.sqn} ;∅) = ∅, valid-peb(apeb)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 7:
m < # apebs, valid-peb(v, m0, apebs[m]), n ≤ (valid-sqns(v, m0, apebs) \ (valid-peb(v, m0, apebs[m]) ⊃ {apebs[m].vidhdr.sqn} ;∅)).max, ∀ v, m0. valid-sqns(v, m0, apebs) ≠ ∅ → valid-sqns(v, m0, apebs).max < n ⊦ valid-sqns(v, m0, apebs) \ {apebs[m].vidhdr.sqn} = ∅, valid-peb(apeb)
Proof step 11:
m < # apebs, n ≤ (valid-sqns(v, m0, apebs) \ (valid-peb(v, m0, apebs[m]) ⊃ {apebs[m].vidhdr.sqn} ;∅)).max, ∀ v, m0. valid-sqns(v, m0, apebs) ≠ ∅ → valid-sqns(v, m0, apebs).max < n ⊦ valid-sqns(v, m0, apebs) \ ∅ = ∅, valid-peb(apeb), valid-peb(v, m0, apebs[m])

Rule argument

pos: right 1

Proof step 6 from proof valid-global-sqn-invalidate (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information