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Proof step 5 from proof toSet-diff-old (previous step)

⊦ a0 ∈ (s0 ++ a) \ x0.toSet ↔ a0 = a ∨ a0 ∈ s0 \ x0.toSet, a ∈ x0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2
Proof step 7:
⊦ a0 ∈ s0 ++ a ∧ ¬ a0 ∈ x0.toSet ↔ a0 = a ∨ a0 ∈ s0 \ x0.toSet, a ∈ x0

Rule argument

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

Additional information

Used terms: x0.toSet, s0 ++ a, a0