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Proof step 8 from proof uni-inj (previous step)

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])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
H ⊥ H1, H ⊥ H0, ∀ r0. (¬ r0 ∈ H → (r0 ∈ H0 ↔ r0 ∈ H1)) ∧ (r0 ∈ H ∨ r0 ∈ H0 → (r0 ∈ H ⊃ H[r0];H0[r0]) = (H ∪ H1)[r0]) ⊦ (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])

Rule argument

Additional information

Used simplifier rules: in-uni: ⊦ H0 ⊥ H1 → (r ∈ H0 ∪ H1 ↔ r ∈ H0 ∨ r ∈ H1);%at-uni: r ∈ H0 ∪ H1 ⊦ H0 ⊥ H1 → (H0 ∪ H1)[r] = (r ∈ H0 ⊃ H0[r];H1[r]);