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Proof step 7 from proof toSet-diff-old (previous step)

⊦ a0 ∈ s0 ++ a ∧ ¬ a0 ∈ x0.toSet ↔ a0 = a ∨ a0 ∈ s0 \ x0.toSet, a ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ a ∈ x.toSet ↔ a ∈ x
Proof step 9:
a0 = a ∨ a0 ∈ s0 \ x0.toSet ⊦ a ∈ x0, (a0 = a ∨ a0 ∈ s0) ∧ ¬ a0 ∈ x0

Rule argument

Additional information

Used simplifier rules: toSet-ext: ⊦ a ∈ x.toSet ↔ a ∈ x;% ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (s)% ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2; (ws)% ⊦ a ≠ b → (a ∈ s ++ b ↔ a ∈ s); (s)