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Proof step 10 from proof sw-del-add (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 11:
r1 ∈ H ⊦ (r1 = r ⊃ o;((H -- r0) ++ r)[r1]) = ((H ++ r)[r, o] -- r0)[r1], r = null, r0 = r, r1 = r0, r ∈ H

Rule argument

pos: left 1

Additional information

Used simplifier rules: in-write: ⊦ r ∈ H[r0, o] ↔ r ∈ H;%in-alloc: r0 ≠ null, ¬ r0 ∈ H ⊦ r ∈ H ++ r0 ↔ r = r0 ∨ r ∈ H;%at-del: r ∈ H ⊦ r ≠ r0 → (H -- r0)[r] = H[r];%at-write: ⊦ H[r0, o][r] = (r = r0 ∧ r0 ∈ H ⊃ o;H[r]);