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Proof step 3 from proof size-one-02 (previous step)

# s = 1, a ∈ s, b ∈ s ⊦ a = b

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
#(s0 ++ a0) = 1, b ∈ s0 ++ a0, a ∈ s0 ++ a0, ∀ a, b. # s0 = 1 ∧ a ∈ s0 ∧ b ∈ s0 → a = b ⊦ a = b

Rule argument

term: s

Additional information

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