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Proof step 11 from proof toSet-diff-old (previous step)

a0 = a ∨ a0 ∈ s0 ∧ ¬ a0 ∈ x0.toSet ⊦ a ∈ x0, (a0 = a ∨ a0 ∈ s0) ∧ ¬ a0 ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
⊦ a ∈ x.toSet ↔ a ∈ x
Proof step 13:
a0 = a ∨ a0 ∈ s0 ∧ ¬ a0 ∈ x0 ⊦ a ∈ x0, (a0 = a ∨ a0 ∈ s0) ∧ ¬ a0 ∈ x0

Rule argument

Additional information

Used simplifier rules: toSet-ext: ⊦ a ∈ x.toSet ↔ a ∈ x;