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Proof step 5 from proof del-new (previous step)

⊦ r = null, r ∈ H, ∀ r0. (r0 ∈ (H ++ r) -- r ↔ r0 ∈ H) ∧ (r0 ∈ (H ++ r) -- r → ((H ++ r) -- r)[r0] = H[r0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
r0 ∈ H ⊦ ((H ++ r) -- r)[r0] = H[r0], r = null, r ∈ H

Rule argument

Additional information

Used simplifier rules: in-del: ⊦ r ∈ H -- r0 ↔ r ≠ r0 ∧ r ∈ H;%in-alloc: r0 ≠ null, ¬ r0 ∈ H ⊦ r ∈ H ++ r0 ↔ r = r0 ∨ r ∈ H;