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Proof step 7 from proof diff-del (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2

Rule argument

spec: <toplevel> name: Difference seq: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2 subst: [s2, s1, a] → [s2, s1, a0] flags: <all paths>

Additional information

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