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

H0 ⊥ H1 ⊦ ∀ r0. (r0 ∈ (H0 ∪ H1) -- r ↔ r0 ∈ (H0 -- r) ∪ (H1 -- r)) ∧ (r0 ∈ (H0 ∪ H1) -- r → ((H0 ∪ H1) -- r)[r0] = ((H0 -- r) ∪ (H1 -- r))[r0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: in-del: ⊦ r ∈ H -- r0 ↔ r ≠ r0 ∧ r ∈ H;%in-uni: ⊦ H0 ⊥ H1 → (r ∈ H0 ∪ H1 ↔ r ∈ H0 ∨ r ∈ H1);%at-del: r ∈ H ⊦ r ≠ r0 → (H -- r0)[r] = H[r];%at-uni: r ∈ H0 ∪ H1 ⊦ H0 ⊥ H1 → (H0 ∪ H1)[r] = (r ∈ H0 ⊃ H0[r];H1[r]);