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Proof step 9 from proof toSet-diff-old (previous step)

a0 = a ∨ a0 ∈ s0 \ x0.toSet ⊦ a ∈ x0, (a0 = a ∨ a0 ∈ s0) ∧ ¬ a0 ∈ x0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2
Proof step 11:
a0 = a ∨ a0 ∈ s0 ∧ ¬ a0 ∈ x0.toSet ⊦ a ∈ x0, (a0 = a ∨ a0 ∈ s0) ∧ ¬ a0 ∈ x0

Rule argument

spec: set-union name: Difference seq: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2 subst: [s2, s1, a] → [x0.toSet, s0, a0] flags: <all paths>

Additional information

Used terms: x0.toSet, s0, a0