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Proof step 6 from proof uni-inj (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r]))
Proof step 8:
H ⊥ H0, H ⊥ H1, ∀ r. (r ∈ H ∪ H0 ↔ r ∈ H ∪ H1) ∧ (r ∈ H ∪ H0 → (H ∪ H0)[r] = (H ∪ H1)[r]) ⊦ (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])

Rule argument

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

Additional information

Used terms: H ∪ H1, H ∪ H0