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Proof step 24 from proof precise-disj (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
H ∪ H5 ∪ H0 = H2 ∪ H3 ∪ H1, P0(H), P0(H3), P1(H2), P1(H5), precise(P1), H2 ⊥ H3, H2 ⊥ H1, H5 ⊥ H0, H3 ⊥ H1, H ⊥ H5, H ⊥ H0, ∀ H, H0, H1, H2. P0(H) ∧ P0(H0) ∧ H ⊥ H1 ∧ H0 ⊥ H2 ∧ H ∪ H1 = H0 ∪ H2 → H = H0 ⊦ H = H3, H2 ∪ H3 ∪ H1 = H3 ∪ H2 ∪ H1, H ⊥ H2

Rule argument

Additional information

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