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Proof step 11 from proof precise-sep (previous step)

H2 ∪ H0 = (H4 ∪ H3) ∪ H, P0(H4), P1(H3), precise(P0), precise(P1), H ⊥ H3, H ⊥ H4, H2 ⊥ H0, H4 ⊥ H3, ∃ H0, H1. H0 ⊥ H1 ∧ H2 = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1) ⊦ H2 = H4 ∪ H3

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
(H5 ∪ H1) ∪ H0 = (H4 ∪ H3) ∪ H, P0(H4), P0(H5), P1(H3), P1(H1), precise(P1), precise(P0), H ⊥ H4, H ⊥ H3, H5 ⊥ H1, H4 ⊥ H3, H0 ⊥ H5, H0 ⊥ H1 ⊦ H5 = H4 ∧ H1 = H3

Rule argument

Additional information

Used simplifier rules: precise-same: precise(P) ⊦ P(H0) ∧ P(H1) ∧ H0 ⊥ H ∧ H1 ⊥ H2 → (H0 ∪ H = H1 ∪ H2 ↔ H0 = H1 ∧ H = H2);% ⊦ H1 ⊥ H2 → (H1 ∪ H2 ⊥ H0 ↔ H0 ⊥ H1 ∧ H0 ⊥ H2); (ws)