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Proof step 6 from proof sw-del-add (previous step)

⊦ r = null, r0 = r, r ∈ H, (r1 = r ∨ r1 ≠ r0 ∧ r1 ∈ H → r1 ≠ r0 ∧ (r1 = r ∨ r1 ∈ H)) ∧ (r1 = r ∨ r1 ≠ r0 ∧ r1 ∈ H → (r1 = r ⊃ o;((H -- r0) ++ r)[r1]) = ((H ++ r)[r, o] -- r0)[r1])

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 4

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