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Proof step 11 from proof adjunction (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 12:
P0(H1), P1(H0), H1 ⊥ H0, H0 ⊥ H1 ∧ P1(H0) → P2(H0 ∪ H1), ∀ H0. H0 ⊥ H1 ∧ P1(H0) → P2(H0 ∪ H1) ⊦ P2(H1 ∪ H0)

Rule argument

pos: left 4 terms: H0 abort: F discard: F computed: F

Additional information

Used terms: H0