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Theorem leninv-succ
⊦ stutf(nσ) → (leninv(nσ) = ⌜n⌝ + ⌜1⌝ ↔ (# nσ = ∞ ⊃ ∃ m. nσ[m] = n ∧ (∀ m0. m < m0 → nσ[m0] = n + 1);nσ[(# nσ).n - 1] = n))
Type: A user defined theorem
Validity: The proof is valid.
Features:
Comment:
Used by:
eq-leninv
,
inv-get-inverse
,
inv-get-inverse-between
,
inv-get-inverse-last
,
inv-get-inverse-succ
,
inv-get-lslen
,
inv-inverts
,
inv-inverts-leq
,
leninv-invlast
,
leninv-invlast-inv
,
leq-leninv
,
less-leninv
,
lsleninv-1
,
Used by units in proved state:
Proof exists?: Yes
View
the proof
Infos about the proof
Proved?: Yes
Interactions: 42
Proof steps: 97
Used lemmas:
stutf-nonempty
,
stutf-def
,
stutf-alltaken
,
stutf-monle
,
monle-leq
,
maxinf-fin
,
leninv-def
,
Used spec theorems:
nat-inf
/
add-fin
nat-inf-basic
/
notinf-fin
Used simplifier rules:
nat-basic1
/
lf-02
nat
/
lep
nat
/
anti
nat
/
eq
nat
/
lels-02
nat
/
zero-01
nat
/
elim-sub
nat
/
eq-01
nat-basic2
/
eq-01
nat
/
succ-01
nat-basic2
/
one-03
nat-basic1
/
one
nat
/
cancel-01
nat
/
fle
nat
/
lels-03
nat
/
le-ls-simp
nat-inf-basic
/
disj
nat-inf
/
addeqinf-cancel
nat-inf
/
inf-less
nat-basic2
/
l-02
nat-basic2
/
l-01
nat
/
fle-01
nat-basic1
/
ls-02
nat-basic2
/
eq
nat-basic2
/
succ
nat-basic1
/
trans-1
nat-basic1
/
lf
nat
/
lels-01
nat-inf
/
less-inf2
nat
/
ler
nat-basic1
/
lf-01
nat
/
eqle-one
nat-inf-basic
/
sel
nat-inf
/
less-fin
stream
/natstream/
size-empty-equiv
nat
/
elim-pred-c
nat-inf
/
add-fin
nat-basic2
/
l
nat-basic2
/
add-base
nat
/
add-01
nat-basic2
/
oz
nat-basic1
/
neq-01
nat
/
lels
nat-basic1
/
f
nat-inf
/
add-assoc
nat-inf
/
add-comm
nat-inf
/
inf-inseq-reduce
nat-basic2
/
s
nat-basic1
/
ref-1
nat-basic2
/
a
nat-basic2
/
c
nat
/
ls-le-simp