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Proof step 5 from proof size (previous step)

y ⊆ x, x ⊆ y ⊦ # x = # y, a ∈ x ↔ a ∈ y, dups(x), dups(y)

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

Additional information

Used simplifier rules: sub-notin: ⊦ x ⊆ y ∧ ¬ a ∈ y → ¬ a ∈ x;%sub-in: ⊦ x ⊆ y ∧ a ∈ x → a ∈ y;%sub-notin: ⊦ x ⊆ y ∧ ¬ a ∈ y → ¬ a ∈ x;%sub-in: ⊦ x ⊆ y ∧ a ∈ x → a ∈ y;