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Proof step 12 from proof key-size (previous step)

a ∈ st1, a = a ∨ a ∈ st1, a ∈ st0, ∀ a0. a0 ∈ st1[a, d] ↔ a0 ∈ st0, ∀ st0. (∀ a. a ∈ st1 ↔ a ∈ st0) → # st1 = # st0 ⊦ # st1 = # st0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information