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Proof step 8 from proof true-elims (previous step)

P(H1 ∪ H0), P0(H2 ∪ H0), H1 ⊥ H0, H2 ⊥ H1, H2 ⊥ H0 ⊦ ∃ H5, H4, H3. H5 ⊥ H3 ∧ H5 ⊥ H4 ∧ H4 ⊥ H3 ∧ H2 ∪ H1 ∪ H0 = H5 ∪ H4 ∪ H3 ∧ true(H5 ∪ H3) ∧ P(H4 ∪ H3)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
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)

Rule argument

Additional information

Used simplifier rules: trueH: ⊦ true(H);