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Proof step 20 from proof size-two (previous step)

s0 ++ a = ∅ ++ a0 ++ b ⊦ 1 = # s0, a0 = b, a ∈ s0

User applied rule insert lemma

And obtained the following new goals

Proof step 21:
⊦ # s = 1 ↔ (∃ a. s = ∅ ++ a)
Proof step 22:
s0 ++ a = ∅ ++ a0 ++ b, # s0 = 1 ↔ (∃ a. s0 = ∅ ++ a) ⊦ 1 = # s0, a0 = b, a ∈ s0

Rule argument

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

Additional information

Used terms: s0