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Proof step 6 from proof del-new (previous step)

r0 ∈ H ⊦ ((H ++ r) -- r)[r0] = H[r0], r = null, r ∈ H

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
r ∈ H ⊦ r ≠ r0 → (H -- r0)[r] = H[r]
Proof step 8:
r0 ∈ H, r0 ∈ H ++ r ⊦ (H ++ r)[r0] = H[r0], r = null, r0 = r, r ∈ H

Rule argument

spec: <toplevel> name: at-del seq: r ∈ H ⊦ r ≠ r0 → (H -- r0)[r] = H[r] subst: [r, r0, H] → [r0, r, H ++ r] flags: <all paths>

Additional information

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