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Proof step 14 from proof uni-ass (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
r ∈ (H0 ∪ H1) ∪ H, H ⊥ H0, H1 ⊥ H, H0 ⊥ H1, r ∈ H0 ∨ r ∈ H1 ∨ r ∈ H ⊦ (r ∈ H0 ∨ r ∈ H1 ⊃ (r ∈ H0 ⊃ H0[r];H1[r]);H[r]) = (H0 ∪ H1 ∪ H)[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]);%disj-distr-uni-com: ⊦ H1 ⊥ H2 → (H1 ∪ H2 ⊥ H0 ↔ H0 ⊥ H1 ∧ H0 ⊥ H2);%disj-com-rw: ⊦ H0 ⊥ H1 → H1 ⊥ H0;