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Proof step 11 from proof uni-ass (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
H ⊥ H0, H1 ⊥ H, H0 ⊥ H1, r ∈ H0 ∨ r ∈ H1 ∨ r ∈ H ⊦ ((H0 ∪ H1) ∪ H)[r] = (H0 ∪ H1 ∪ H)[r]

Rule argument

Additional information

Used simplifier rules: disj-distr-uni: ⊦ H1 ⊥ H → (H0 ⊥ H1 ∪ H ↔ H0 ⊥ H1 ∧ H0 ⊥ H);%disj-com-rw: ⊦ H0 ⊥ H1 → H1 ⊥ H0;%in-uni: ⊦ H0 ⊥ H1 → (r ∈ H0 ∪ H1 ↔ r ∈ H0 ∨ r ∈ H1);