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Proof step 5 from proof uni-inj (previous step)

H ∪ H0 = H ∪ H1, H ⊥ H0, 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 6:
H ∪ H0 = H ∪ H1, H ⊥ H1, H ⊥ H0 ⊦ (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])

Rule argument

Additional information