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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
H0 ⊆ H1, H1 \ H0 ⊥ H0, r ∈ H1 ∨ r ∈ H0 ⊦ (r ∈ H1 ∧ ¬ r ∈ H0 ⊃ H1[r];H0[r]) = H1[r]

Rule argument

Additional information

Used simplifier rules: diff-in: ⊦ r ∈ H0 \ H1 ↔ r ∈ H0 ∧ ¬ r ∈ H1;%diff-at: r ∈ H0, ¬ r ∈ H1 ⊦ (H0 \ H1)[r] = H0[r];%in-uni: ⊦ H0 ⊥ H1 → (r ∈ H0 ∪ H1 ↔ r ∈ H0 ∨ r ∈ H1);