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Proof step 7 from proof subsize-eq (previous step)

# s = # s0, a ∈ s → a ∈ s0, ∀ a. a ∈ s → a ∈ s0, a ∈ s0 ⊦ a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
# s = # s0, a ∈ s0, ∀ a. a ∈ s → a ∈ s0 ⊦ a ∈ s

Rule argument

Additional information