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Proof step 12 from proof uni-inj (previous step)

(r ∈ H ⊃ H[r];H0[r]) = (H ∪ H1)[r], H ⊥ H0, H ⊥ H1, ¬ r ∈ H → (r ∈ H0 ↔ r ∈ H1), ∀ r0. (¬ r0 ∈ H → (r0 ∈ H0 ↔ r0 ∈ H1)) ∧ (r0 ∈ H ∨ r0 ∈ H0 → (r0 ∈ H ⊃ H[r0];H0[r0]) = (H ∪ H1)[r0]) ⊦ (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

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-l: ⊦ r ∈ H0 ∧ H0 ⊥ H1 → (r ∈ H1 ↔ false);%in-uni: ⊦ H0 ⊥ H1 → (r ∈ H0 ∪ H1 ↔ r ∈ H0 ∨ r ∈ H1);%disj-r: ⊦ r ∈ H1 ∧ H0 ⊥ H1 → (r ∈ H0 ↔ false);%at-uni: r ∈ H0 ∪ H1 ⊦ H0 ⊥ H1 → (H0 ∪ H1)[r] = (r ∈ H0 ⊃ H0[r];H1[r]);