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Proof step 2 from proof diff-union-02 (previous step)

⊦ a ∈ (s ∪ s2) \ s0 ↔ a ∈ s2 \ s0, a ∈ s, a ∈ s1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2

Rule argument

spec: inst: name: Difference seq: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2 subst: [s2, s1, a] → [s0, s ∪ s2, a] flags: rewrite, current

Additional information

Used simplifier rules: unin: ⊦ ¬ a ∈ s → (a ∈ s ∪ s0 ↔ a ∈ s0);%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2; Used terms: s0, s ∪ s2, a