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Proof step 2 from proof diff-diff-right (previous step)

a ∈ s2 ⊦ s2 = ∅, a ∈ s0 \ s1 \ s2 ↔ a ∈ s0, # s2 = 0

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] → [s1 \ s2, s0, a] flags: rewrite, current

Additional information

Used simplifier rules: Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2; Used terms: s1 \ s2, s0, a