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Proof step 1 from proof size-two-03 (conclusion)

1 < # s ⊦ ∃ a, b. a ≠ b ∧ a ∈ s ∧ b ∈ s

User applied rule structural induction

And obtained the following new goals

Proof step 2:
1 < # s0 → (∃ a, b. a ≠ b ∧ a ∈ s0 ∧ b ∈ s0), 1 < #(s0 ++ a) ⊦ ∃ a0, b. a0 ≠ b ∧ a0 ∈ s0 ++ a ∧ b ∈ s0 ++ a

Rule argument

indvar: s

Additional information

Used simplifier rules: gen: ⊦ ;%Size-empty: ⊦ # ∅ = 0;