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Proof step 5 from proof Diff-ext (previous step)

∀ a, st. a ∈ st \ s0 ↔ a ∈ st ∧ ¬ a ∈ s0 ⊦ a ≠ a0 ∧ a ∈ st \ s0 ↔ a ∈ st ∧ a ≠ a0 ∧ ¬ a ∈ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information