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Proof step 3 from proof uni-ass (previous step)

H ⊥ H0, H1 ⊥ H, H0 ⊥ H1 ⊦ (H0 ∪ H1) ∪ H = H0 ∪ H1 ∪ 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:
H0 ⊥ H1, H1 ⊥ H, H ⊥ H0 ⊦ ∀ r. (r ∈ (H0 ∪ H1) ∪ H ↔ r ∈ H0 ∪ H1 ∪ H) ∧ (r ∈ (H0 ∪ H1) ∪ H → ((H0 ∪ H1) ∪ H)[r] = (H0 ∪ H1 ∪ H)[r])

Rule argument

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

Additional information

Used terms: H0 ∪ H1 ∪ H, (H0 ∪ H1) ∪ H