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Proof step 2 from proof toSet-diff-old (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)
Proof step 4:
⊦ a ∈ x0, ∀ a0. a0 ∈ (s0 ++ a) \ x0.toSet ↔ a0 ∈ (s0 \ x0.toSet) ++ a

Rule argument

spec: set-basic name: Extension seq: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2) subst: [s2, s1] → [(s0 \ x0.toSet) ++ a, (s0 ++ a) \ x0.toSet] flags: <all paths>

Additional information

Used terms: (s0 \ x0.toSet) ++ a, (s0 ++ a) \ x0.toSet