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Proof step 4 from proof diff-del (previous step)

a0 ∈ s1, a1 ∈ s1 ⊦ a0 ∈ s2, a ∈ s1 \ s2, a0 ∈ (s1 -- a) \ (s2 -- a)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ 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] → [s2 -- a, s1 -- a, a0] flags: rewrite, current

Additional information

Used simplifier rules: Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;%del-rax: ⊦ a ∈ s -- b ↔ ¬ ¬(a ≠ b ∧ a ∈ s);%Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2; Used terms: s2 -- a, s1 -- a, a0