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

a ∈ s0, s ⊆ s0, # s0 ≤ # s ⊦ a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# s0 ≤ # s, s ⊆ s0, a ∈ s0 ⊦ a ∈ s

Rule argument

Additional information

Used simplifier rules: ne-01: ⊦ a ∈ s → s ≠ ∅;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);