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Proof step 16 from proof subset-is-union (previous step)

s0 ∩ s = ∅ ++ a, a ∈ s ⊦ # s = 0, s = ∅, a ∈ s0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)
Proof step 18:
∀ a0. a0 ∈ s0 ∩ s ↔ a0 ∈ ∅ ++ a, a ∈ s ⊦ # s = 0, s = ∅, a ∈ s0

Rule argument

spec: set-basic inst: name: Extension seq: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2) subst: [s2, s1] → [∅ ++ a, s0 ∩ s] flags: rewrite

Additional information

Used terms: ∅ ++ a, s0 ∩ s