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Proof step 4 from proof size-eq-ext (previous step)

s0 ≠ s1, ∀ a. a ∈ s0 ↔ a ∈ s1 ⊦ # s0 = # s1

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:
∃ a. ¬(a ∈ s0 ↔ a ∈ s1), ∀ a. a ∈ s0 ↔ a ∈ s1 ⊦ # s0 = # s1

Rule argument

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

Additional information

Used terms: s1, s0