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Proof step 3 from proof del-new (previous step)

⊦ (H ++ r) -- r = H, r = null, r ∈ H

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r]))
Proof step 5:
⊦ r = null, r ∈ H, ∀ r0. (r0 ∈ (H ++ r) -- r ↔ r0 ∈ H) ∧ (r0 ∈ (H ++ r) -- r → ((H ++ r) -- r)[r0] = H[r0])

Rule argument

spec: <toplevel> name: ext seq: ⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])) subst: [H1, H0] → [H, (H ++ r) -- r] flags: <all paths>

Additional information

Used terms: H, (H ++ r) -- r