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Proof step 3 from proof int-ins (previous step)

∀ a0. ¬ a0 ∈ s ∨ a0 ≠ a ∧ ¬ a0 ∈ s0 ⊦ (∀ a0. ¬ a0 ∈ s ∨ ¬ a0 ∈ s0) ∧ ¬ a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a0. ¬ a0 ∈ s ∨ a0 ≠ a ∧ ¬ a0 ∈ s0 ⊦ (¬ a0 ∈ s ∨ ¬ a0 ∈ s0) ∧ ¬ a ∈ s

Rule argument

Additional information