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Proof step 13 from proof lebs-del-subseteq (previous step)

@(m, m0, m1) ∈ ri -- key ⊦ m ∈ ns, (ri -- key).lebs ⊆ ri.lebs, ∃ adr. adr.lnum = m ∧ adr ∈ ri

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 3 terms: @(m, m0, m1)

Additional information

Used simplifier rules: ⊦ @(m, m0, m1).lnum = m; (s)% ⊦ ¬ adr ∈ ri → ¬ adr ∈ ri -- key; (s) Used terms: @(m, m0, m1)