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Proof step 5 from proof true-elims (
previous step
)
∃ H0, H1, H2. H0 ⊥ H2 ∧ H0 ⊥ H1 ∧ H1 ⊥ H2 ∧ H = H0 ∪ H1 ∪ H2 ∧ P0(H0 ∪ H2) ∧ P(H1 ∪ H2) ⊦ ((true ⊗ P))(H)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 6
:
P(H1 ∪ H0), P0(H2 ∪ H0), H2 ⊥ H0, H2 ⊥ H1, H1 ⊥ H0 ⊦ ((true ⊗ P))(H2 ∪ H1 ∪ H0)
Rule argument
Additional information