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Theorem abs-nodes-split-h
⊦ abs-nodes(buf, m, x0 + x1, lax, weak) → (∃ buf0, buf1, lax0, lax1. lax = lax0 + lax1 ∧ buf = buf0 + buf1 ∧ abs-nodes(buf0, m, x0, lax0, false) ∧ abs-nodes(buf1, m + # buf0, x1, lax1, weak))
Type: A user defined theorem
Validity: The proof is valid.
Features:
Comment:
Used by:
abs-nodes-split
,
Used by units in proved state:
Proof exists?: Yes
View
the proof
Infos about the proof
Proved?: Yes
Interactions: 10
Proof steps: 24
Used lemmas:
ls-abs-node-buf-size
,
abs-nodes-unique-strong-h
,
abs-nodes-padding-prefix
,
abs-nodes-first-adr-explicit
,
abs-nodes-rec
,
abs-nodes-mkbuf0-empty-h
,
Used spec theorems:
Used simplifier rules:
list
/
Nil
list
/laddress-list/
Nil
array-splice
/buffer/
append-zero-left
list
/
c
laddress
/
elim
list
/laddress-list/
in-rec
list
/
a
nat-basic
/
add-base
abs-node
/
abs-padding-mkbuf0
list
/laddress-list/
in-append
list
/laddress-list/
in-14
list-data
/laddress-list/
elim-list-c
list
/laddress-list/
c
nat
/
lels
nat
/
le-add
abs-node
/
buf-size-zero
array
/buffer/
Size-mkarray
nat-basic
/
z-01
nat-mult
/
two-zero-01
node-header-size
/
nodeheader-notzero
nat-mult
/
mult-a
nat-mult
/
mult-c
nat-basic
/
c
nat-basic
/
a
array-splice
/buffer/
append-len
laddress
/
inj
list
/laddress-list/
app-07
array-splice
/buffer/
append-a
abs-node
/
abs-node-no-abs-padding
list
/laddress-list/
e-01
list
/laddress-list/
a
list
/laddress-list/
c-08
laddress
/
sel-02
laddress
/
sel-01
abs-node
/
abs-node-abs-node
list
/laddress-list/
c-07
array-splice
/buffer/
append-eq-left