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Proof step 2 from proof sub-one (previous step)

# s0 = 1 ⊦ s ⊆ s0 ↔ s = s0, s = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# s0 = 1, ∀ a. a ∈ s → a ∈ s0 ⊦ # s = 0, s = s0, s = ∅

Rule argument

Additional information

Used simplifier rules: Subset: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2);%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0;%size-zero: ⊦ # s ≠ 0 → s ≠ ∅;