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Proof step 3 from proof owns-one-none (previous step)

g(r) = (r = r0 ⊃ a0;g(r0)), r0 ∈ rs, r ∈ rs, ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r) ⊦ # rs = 0, g(r) = a0, r = null, rs = ∅

User applied rule case distinction

And obtained the following new goals

Proof step 4:
g(r) = g(r0), r ∈ rs, r0 ∈ rs, ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r) ⊦ # rs = 0, g(r) = a0, r = null, r = r0, rs = ∅

Rule argument

pos: left 1

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