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Proof step 2 from proof at-uni (previous step)

r ∈ H0 ∪ H1, H0 ⊥ H1 ⊦ (H0 ∪ H1)[r] = (r ∈ H0 ⊃ H0[r];H1[r])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
H0 ⊥ H1, r ∈ H0 ∨ r ∈ H1 ⊦ (H0 ∪ H1)[r] = (r ∈ H0 ⊃ H0[r];H1[r])

Rule argument

Additional information

Used simplifier rules: in-uni: ⊦ H0 ⊥ H1 → (r ∈ H0 ∪ H1 ↔ r ∈ H0 ∨ r ∈ H1);