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

⊦ r = null, r0 = r, r ∈ rs, r ∈ H, (r0 = r ∨ r0 ∈ H ∧ ¬ r0 ∈ rs → (r0 = r ∨ r0 ∈ H) ∧ ¬ r0 ∈ rs) ∧ ((r0 = r ∨ r0 ∈ H) ∧ ¬ r0 ∈ rs → ((H ++ r)[r, o] \ rs)[r0] = (r0 = r ⊃ o;((H \ rs) ++ r)[r0]))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
r0 ∈ H ⊦ ((H ++ r)[r, o] \ rs)[r0] = H[r0], r = null, r ∈ H, r ∈ rs, r0 ∈ rs

Rule argument

Additional information

Used simplifier rules: in-diffs: ⊦ r ∈ H \ rs ↔ r ∈ H ∧ ¬ r ∈ rs;%at-alloc: r ∈ H, r0 ≠ null, ¬ r0 ∈ H ⊦ r ≠ r0 → (H ++ r0)[r] = H[r];%at-diffs: ⊦ r ∈ H ∧ ¬ r ∈ rs → (H \ rs)[r] = H[r];