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Proof step 1 from proof sub-one (conclusion)

⊦ s ≠ ∅ ∧ # s0 = 1 → (s ⊆ s0 ↔ s = s0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
# s0 = 1, s ⊆ s0 ⊦ # s = 0, s = s0, s = ∅

Rule argument

Additional information

Used simplifier rules: sub-idem: ⊦ s ⊆ s;%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0;%size-zero: ⊦ # s ≠ 0 → s ≠ ∅;