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Proof step 10 from proof size-two (previous step)

1 = # s0 ⊦ a ∈ s0, ∃ a0, b. a0 ≠ b ∧ s0 ++ a = ∅ ++ a0 ++ b

User applied rule insert lemma

And obtained the following new goals

Proof step 11:
⊦ # s = 1 → (∃ a. s = ∅ ++ a)
Proof step 12:
# s0 = 1 → (∃ a. s0 = ∅ ++ a), 1 = # s0 ⊦ a ∈ s0, ∃ a0, b. a0 ≠ b ∧ s0 ++ a = ∅ ++ a0 ++ b

Rule argument

spec: set-basic inst: name: size-one seq: ⊦ # s = 1 → (∃ a. s = ∅ ++ a) subst: [s] → [s0] flags: current, []

Additional information

Used terms: s0