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Proof step 8 from proof precise-sep (previous step)

H2 ∪ H0 = H1 ∪ H, ((P0 * P1))(H2), precise(P0), precise(P1), H1 ⊥ H, H2 ⊥ H0, ∃ H0, H2. H0 ⊥ H2 ∧ H1 = H0 ∪ H2 ∧ P0(H0) ∧ P1(H2) ⊦ H2 = H1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
H2 ∪ H0 = (H4 ∪ H3) ∪ H, ((P0 * P1))(H2), P0(H4), P1(H3), precise(P1), precise(P0), H4 ⊥ H3, H2 ⊥ H0, H ⊥ H4, H ⊥ H3 ⊦ H2 = H4 ∪ H3

Rule argument

Additional information

Used simplifier rules: ⊦ H1 ⊥ H2 → (H1 ∪ H2 ⊥ H0 ↔ H0 ⊥ H1 ∧ H0 ⊥ H2); (ws)