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Proof step 6 from proof sw-put (previous step)

r0 ∈ H, r1 ∈ H, r ∈ H ⊦ (r = r1 ⊃ o1;(r = r0 ⊃ o0;H[r])) = (r = r0 ⊃ o0;(r = r1 ⊃ o1;H[r])), r0 = r1

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

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