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Proof step 9 from proof true-elims (previous step)

P(H1 ∪ H0), P0(H2 ∪ H0), H2 ⊥ H0, H2 ⊥ H1, H1 ⊥ H0 ⊦ ∃ H, H3, H4. H ⊥ H4 ∧ H ⊥ H3 ∧ H3 ⊥ H4 ∧ H2 ∪ H1 ∪ H0 = H ∪ H3 ∪ H4 ∧ P(H3 ∪ H4)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: H2, H1, H0

Additional information

Used terms: H2, H1, H0