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Proof step 2 from proof subseteq-inter-empty-empty (previous step)

s1 ∩ s2 = ∅, s0 ⊆ s1 ⊦ #(s0 ∩ s2) = 0, s0 ∩ s2 = ∅

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:
s1 ∩ s2 = ∅, ∀ a. a ∈ s0 → a ∈ s1 ⊦ #(s0 ∩ s2) = 0, s0 ∩ s2 = ∅

Rule argument

spec: set-basic inst: name: Subset seq: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2) subst: [s2, s1] → [s1, s0] flags: rewrite

Additional information

Used terms: s1, s0