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Theorem sharing
⊦ ((P0 ⊗ P1))(H) ↔ (∃ H0, H1, H2. H0 ⊥ H2 ∧ H0 ⊥ H1 ∧ H1 ⊥ H2 ∧ H = H0 ∪ H1 ∪ H2 ∧ P0(H0 ∪ H2) ∧ P1(H1 ∪ H2))
Type: An axiom
Validity: The proof is valid.
Features:
Comment:
Used by:
introduce-sharing
,
sharing-com
,
sharing-emp
,
true-elims
,
Used by units in proved state:
Proof exists?: No