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Proof step 12 from proof adjunction (previous step)

P0(H1), P1(H0), H1 ⊥ H0, H0 ⊥ H1 ∧ P1(H0) → P2(H0 ∪ H1), ∀ H0. H0 ⊥ H1 ∧ P1(H0) → P2(H0 ∪ H1) ⊦ P2(H1 ∪ H0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
P0(H1), P1(H0), P2(H0 ∪ H1), H1 ⊥ H0, ∀ H0. H0 ⊥ H1 ∧ P1(H0) → P2(H0 ∪ H1) ⊦ P2(H1 ∪ H0)

Rule argument

Additional information

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