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Proof step 14 from proof lebs-lookup-subseteq (previous step)

@(m, m0, m1) ∈ ri[ks] ⊦ # ri[ks] = 0, ri[ks] = ∅, m ∈ ns, ri[ks].lebs ⊆ ri.lebs, @(m, m0, m1).lnum = m ∧ @(m, m0, m1) ∈ ri

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
@(m, m0, m1) ∈ ri[ks] ⊦ # ri[ks] = 0, ri[ks] = ∅, m ∈ ns, @(m, m0, m1) ∈ ri, ri[ks].lebs ⊆ ri.lebs

Rule argument

Additional information

Used simplifier rules: ⊦ @(m, m0, m1).lnum = m; (s)