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Proof step 39 from proof mapex-new (previous step)

P(H0), r0 ∈ (∅ ++ r)[r, o] ∪ H0, r0 ≠ r → (r0 ∈ H ↔ r0 ∈ H0), r0 = r ∨ r0 ∈ H → (H ++ r)[r0] = (r0 = r ⊃ o;H0[r0]) ⊦ r = null, P(H), r ∈ H, r ∈ H0, (r0 ∈ H0 ↔ r0 ∈ H) ∧ (r0 ∈ H0 → H0[r0] = H[r0])

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 4

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