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Proof step 4 from proof sub-restr-sub (previous step)

⊦ ((ns -- n) \ ns0) -- n = (ns \ ns0) -- n, ∀ n0. n0 ∈ (ns -- n) \ ns0 ↔ n0 ∈ ns \ ns0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ ((ns -- n) \ ns0) -- n = (ns \ ns0) -- n, n0 ∈ (ns -- n) \ ns0 ↔ n0 ∈ ns \ ns0

Rule argument

Additional information