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Proof step 2 from proof union-left-singleton (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r]))
Proof step 4:
r ∈ H ⊦ r = null, ∀ r0. (r0 ∈ (∅ ++ r)[r, H[r]] ∪ (H -- r) ↔ r0 ∈ H) ∧ (r0 ∈ (∅ ++ r)[r, H[r]] ∪ (H -- r) → ((∅ ++ r)[r, H[r]] ∪ (H -- 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, (∅ ++ r)[r, H[r]] ∪ (H -- r)] flags: <all paths>

Additional information

Used terms: H, (∅ ++ r)[r, H[r]] ∪ (H -- r)