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Proof step 26 from proof precise-sep (previous step)

H5 ∪ H1 ∪ H0 = (H4 ∪ H3) ∪ H, P0(H5), P0(H4), P1(H1), P1(H3), H0 ⊥ H1, H4 ⊥ H3, H1 ⊥ H5, H ⊥ H3, H4 ⊥ H, H5 ⊥ H1, H1 ⊥ H0, H0 ⊥ H5, H5 ⊥ H1 ∪ H0 ∧ H4 ⊥ H3 ∪ H ∧ H5 ∪ H1 ∪ H0 = H4 ∪ H3 ∪ H → H5 = H4, ∀ H, H0, H1, H2. P1(H) ∧ P1(H0) ∧ H ⊥ H1 ∧ H0 ⊥ H2 ∧ H ∪ H1 = H0 ∪ H2 → H = H0 ⊦ H5 = H4 ∧ H1 = H3

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
H5 ∪ H1 ∪ H0 = (H4 ∪ H3) ∪ H, P0(H4), P0(H5), P1(H3), P1(H1), H0 ⊥ H5, H4 ⊥ H, H ⊥ H3, H1 ⊥ H5, H4 ⊥ H3, H0 ⊥ H1, H5 ⊥ H1 ∪ H0 ∧ H4 ⊥ H3 ∪ H ∧ (H4 ∪ H3) ∪ H = H4 ∪ H3 ∪ H → H5 = H4, ∀ H, H0, H1, H2. P1(H) ∧ P1(H0) ∧ H ⊥ H1 ∧ H0 ⊥ H2 ∧ H ∪ H1 = H0 ∪ H2 → H = H0 ⊦ H5 = H4 ∧ H1 = H3

Rule argument

Additional information

Used simplifier rules: ⊦ H0 ⊥ H1 → H1 ⊥ H0; (s)