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Theorem precise
⊦ precise(hP) ↔ (∀ h, h0, h1, h2. hP(h) ∧ hP(h0) ∧ h ⊥ h1 ∧ h0 ⊥ h2 ∧ h ∪ h1 = h0 ∪ h2 → h = h0)
Type: An axiom
Validity: The proof is valid.
Features:
Comment:
Used by:
precise-alt-def
,
precise-and
,
precise-disj
,
precise-emp
,
precise-same
,
precise-sep
,
precise-split-l
,
precise-this
,
Used by units in proved state:
linked-list
,
maplet
,
Proof exists?: No