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Proof step 3 from proof rem-union-ins (previous step)

a ∈ s1 ⊦ # s1 = 0, (s0 ∪ (s1 -- a)) ++ a = s0 ∪ s1, s1 = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)

Rule argument

spec: set-basic name: Extension seq: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2) subst: [s2, s1] → [s0 ∪ s1, (s0 ∪ (s1 -- a)) ++ a] flags: <all paths>

Additional information

Used simplifier rules: una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%unindel: ⊦ ¬ a ∈ s ∪ s0 → ¬ a ∈ (s -- b) ∪ s0;%unrax: ⊦ a ∈ s ∪ s0 ↔ a ∈ s ∨ a ∈ s0;%Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;% ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (s) Used terms: s0 ∪ s1, (s0 ∪ (s1 -- a)) ++ a