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Proof step 8 from proof sw-alloc-put (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 9:
r = r1, r ∈ H, r1 ∈ H ⊦ o1 = (H[r1, o1] ++ r0)[r], r0 = null, r0 ∈ H
Proof step 14:
r ∈ H, r1 ∈ H ⊦ (H ++ r0)[r] = (H[r1, o1] ++ r0)[r], r = r1, r0 = null, r0 ∈ H

Rule argument

pos: right 1

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