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Proof step 5 from proof sub-size-eq (previous step)

s ⊆ s0, # s0 ≤ # s ⊦ a ∈ s ↔ a ∈ s0

User applied rule case distinction

And obtained the following new goals

Proof step 6:
a ∈ s0, s ⊆ s0, # s0 ≤ # s ⊦ a ∈ s

Rule argument

pos: right 1

Additional information

Used simplifier rules: sub-in: ⊦ a ∈ s ∧ s ⊆ s0 → a ∈ s0;%ne-01: ⊦ a ∈ s → s ≠ ∅;