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Proof step 1 from proof two-01 (conclusion)
a ∈ s, s ≠ ∅ ++ a ⊦ ∃ b. b ∈ s ∧ b ≠ a
System applied rule normalize
And obtained the following new goals
Proof step 2
:
a ∈ s ⊦ s = ∅ ++ a, ∃ b. b ∈ s ∧ b ≠ a
Rule argument
Additional information