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Proof step 5 from proof split-union (previous step)

P0(H0), P1(H1), H0 ⊥ H1 ⊦ ∃ H2, H3. H2 ⊥ H3 ∧ H0 ∪ H1 = H2 ∪ H3 ∧ P0(H2) ∧ P1(H3)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: H0, H1

Additional information

Used terms: H0, H1