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Proof step 4 from proof subsize-eq (previous step)

# s = # s0, s ⊆ s0, a ∈ s0 ⊦ a ∈ s

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2)
Proof step 6:
# s = # s0, ∀ a. a ∈ s → a ∈ s0, a ∈ s0 ⊦ a ∈ s

Rule argument

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

Additional information

Used terms: s0, s