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Proof step 6 from proof size-eq-ext (previous step)

∃ a. ¬(a ∈ s0 ↔ a ∈ s1), ∀ a. a ∈ s0 ↔ a ∈ s1 ⊦ # s0 = # s1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ a. a ∈ s0 ↔ a ∈ s1 ⊦ # s0 = # s1, a ∈ s0 ↔ a ∈ s1

Rule argument

Additional information