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Proof step 2 from proof size-eq-ext (previous step)

∀ a. a ∈ s0 ↔ a ∈ s1 ⊦ # s0 = # s1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ s0 = s1 → (# s0 ≠ # s1 ↔ false)
Proof step 4:
s0 ≠ s1, ∀ a. a ∈ s0 ↔ a ∈ s1 ⊦ # s0 = # s1

Rule argument

spec: inst: name: size-eq seq: ⊦ s0 = s1 → (# s0 ≠ # s1 ↔ false) subst: [s1, s0] → [s1, s0] flags: rewrite, current

Additional information

Used terms: s1, s0