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Proof step 1 from proof equal-add-in (conclusion)

⊦ s ++ a = s0 ++ b ∧ s0 ++ b = (s1 -- c) ++ a0 ∧ b ≠ a0 → b ∈ s1 -- c

System applied rule normalize

And obtained the following new goals

Proof step 2:
s ++ a = s0 ++ b, s0 ++ b = (s1 -- c) ++ a0 ⊦ b = a0, b ∈ s1 -- c

Rule argument

Additional information