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Proof step 7 from proof precise-split-l (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
H0 ∪ H1 = H2 ∪ H, P0(H2), P0(H0), P1(H), H2 ⊥ H, H0 ⊥ H1, ∀ H, H0, H1, H2. P0(H) ∧ P0(H0) ∧ H ⊥ H1 ∧ H0 ⊥ H2 ∧ H ∪ H1 = H0 ∪ H2 → H = H0 ⊦ P1(H1)

Proof step 7 from proof precise-split-l (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information