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Proof step 1 from proof subsize-eq (conclusion)

⊦ s ⊆ s0 ∧ # s = # s0 → s = s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
# s = # s0, s ⊆ s0 ⊦ a ∈ s ↔ a ∈ s0

Rule argument

Additional information

Used simplifier rules: Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2);