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Proof step 2 from proof ext-rec (previous step)

∀ b. b ∈ s1 ↔ b = a ∨ b ∈ s0 ⊦ s1 = s0 ++ a

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:
∀ b. b ∈ s1 ↔ b = a ∨ b ∈ s0 ⊦ ∀ a0. a0 ∈ s1 ↔ a0 ∈ s0 ++ a

Rule argument

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

Additional information

Used terms: s0 ++ a, s1