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Proof step 14 from proof sw-del-add (previous step)

r1 ∈ H ⊦ ((H -- r0) ++ r)[r1] = (H ++ r)[r1], r = null, r0 = r, r1 = r0, r ∈ H

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
r ∈ H, r0 ≠ null, ¬ r0 ∈ H ⊦ r ≠ r0 → (H ++ r0)[r] = H[r]
Proof step 16:
r1 ∈ H, r1 ∈ H -- r0 ⊦ (H -- r0)[r1] = (H ++ r)[r1], r = null, r0 = r, r1 = r0, r1 = r, r ∈ H, r ∈ H -- r0

Rule argument

spec: <toplevel> name: at-alloc seq: r ∈ H, r0 ≠ null, ¬ r0 ∈ H ⊦ r ≠ r0 → (H ++ r0)[r] = H[r] subst: [r, r0, H] → [r1, r, H -- r0] flags: <all paths>

Additional information

Used simplifier rules: in-del: ⊦ r ∈ H -- r0 ↔ r ≠ r0 ∧ r ∈ H; Used terms: r1, r, H -- r0