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Proof step 3 from proof sub-one (previous step)

# s0 = 1, ∀ a. a ∈ s → a ∈ s0 ⊦ # s = 0, s = s0, s = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ # s = 1 ↔ (∃ a. s = ∅ ++ a)
Proof step 5:
∀ a. a ∈ s → a ∈ s0, ∃ a. s0 = ∅ ++ a ⊦ # s = 0, s = ∅, s = s0

Rule argument

spec: <toplevel> name: size-one-03 seq: ⊦ # s = 1 ↔ (∃ a. s = ∅ ++ a) subst: [s] → [s0] flags: <all paths>

Additional information

Used terms: s0