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Proof step 8 from proof sub-one (previous step)

∀ a0. a0 ∈ s → a0 = a ⊦ # s = 0, s = ∅ ++ a, s = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)
Proof step 10:
∀ a0. a0 ∈ s → a0 = a ⊦ # s = 0, s = ∅, ∀ a0. a0 ∈ s ↔ a0 ∈ ∅ ++ a

Rule argument

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

Additional information

Used terms: ∅ ++ a, s