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Proof step 6 from proof add-inter-rem (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)

Rule argument

spec: set-basic inst: name: Extension seq: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2) subst: [s2, s1] → [s0 ∩ s1, ((s0 ++ a) ∩ s1) -- a] flags: rewrite

Additional information

Used simplifier rules: del-rax: ⊦ a ∈ s -- b ↔ ¬ ¬(a ≠ b ∧ a ∈ s);%int-rax: ⊦ a ∈ s1 ∩ s2 ↔ ¬ ¬(a ∈ s1 ∧ a ∈ s2);%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;%int-c: ⊦ s ∩ s0 = s0 ∩ s;%Intersect: ⊦ a ∈ s1 ∩ s2 ↔ a ∈ s1 ∧ a ∈ s2;%in-02: ⊦ a ≠ b → (a ∈ s ++ b ↔ a ∈ s); from specification set-basic%in: ⊦ a ∈ s → a ∈ s ++ b; from specification set-basic Used terms: s0 ∩ s1, ((s0 ++ a) ∩ s1) -- a