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Theorem abs-nodes-one-h
⊦ abs-nodes(buf, m, a ', lax, weak) ↔ (∃ buf0, buf1. buf = buf0 + buf1 ∧ ( abs-padding(buf0) ∧ abs-nodes(buf1, m + # buf0, a ', lax, weak) ∨ lax = @(m, a.flashsize) ' ∧ abs-node(buf0, a) ∧ abs-nodes(buf1, m + # buf0, [], [], weak)))
Type: A user defined theorem
Validity: The proof is valid.
Features:
Comment:
Used by:
abs-nodes-add-crash-aligned-h
,
abs-nodes-one
,
abs-nodes-subarray-failed
,
Used by units in proved state:
Proof exists?: Yes
View
the proof
Infos about the proof
Proved?: Yes
Interactions: 6
Proof steps: 18
Used lemmas:
abs-nodes-elems-empty-adr-empty-h
,
ls-abs-node-buf-size
,
abs-nodes-rec
,
Used spec theorems:
list
/
One
list
/
c
Used simplifier rules:
encoding-node-base
/
node-flash-size
list
/
a
list
/
app
list
/laddress-list/
e
list-data
/laddress-list/
elim-list-c
list
/laddress-list/
e-01
list
/laddress-list/
c
laddress
/
elim
list
/laddress-list/
app
list
/laddress-list/
c-04
abs-node
/
abs-padding-eq
list
/laddress-list/
c-03
list
/laddress-list/
a
laddress
/
inj
array-splice
/buffer/
append-a
abs-node
/
abs-node-no-abs-padding
abs-node
/
abs-node-abs-node
array-splice
/buffer/
append-eq-left
nat-basic
/
c