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Theorem ext

¬ null ∈ h0, ¬ null ∈ h1 ⊦ h0 = h1 ↔ (∀ r. (r ∈ h0 ↔ r ∈ h1) ∧ (r ∈ h0 → h0[r]h = h1[r]h))

Type: A proof obligation
Validity: The proof is valid.
Features:
Comment: |- h0 = h1 <-> (all r. (r \in h0 <-> r \in h1) and (r \in h0 -> h0(r) = h1(r)))
Used by:
Used by units in proved state:
Proof exists?: Yes

View the proof

Infos about the proof

Proved?: Yes
Interactions: 4
Proof steps: 7
Used lemmas:
Used spec theorems:
- map/Extension
- map/notmemlookup
Used simplifier rules:
- heapmap/get-def