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Proof step 4 from proof del-e (previous step)

a = b ⊦ (∀ a0. a0 = b ∨ ¬ a0 ∈ s ++ a) ↔ (∀ a0. ¬ a0 ∈ s) ∧ a = b, a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ (∀ a0. a0 = b ∨ ¬ a0 ∈ s ++ b) ↔ (∀ a0. ¬ a0 ∈ s), b ∈ s

Rule argument

Additional information