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Proof step 5 from proof elems-in-fwd (previous step)

n ∈ nx, ∀ n. n ∈ elems nx → n ∈ ns ⊦ n ∈ ns

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n ∈ nx, ∀ n0. n0 ∈ nx → n0 ∈ ns ⊦ n ∈ ns

Rule argument

Additional information

Used simplifier rules: elems-def: ⊦ n ∈ elems nx ↔ n ∈ nx;