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Theorem one-oc
⊦ #oc(n, A, m0) = 1 ↔ (∃ m. m < m0 ∧ A[m] = n ∧ (∀ m1. m1 ≠ m ∧ m1 < m0 → A[m1] ≠ n))
Type: A user defined theorem
Validity: The proof is valid.
Features:
Comment:
Used by:
nodups-numoc
,
Used by units in proved state:
Proof exists?: Yes
View
the proof
Infos about the proof
Proved?: Yes
Interactions: 16
Proof steps: 49
Used lemmas:
no-oc
,
Used spec theorems:
Used simplifier rules:
nat-basic1
/
zero-01
array-perm
/array_of_nat/
numoc-base
nat-basic2
/
s
array-perm
/array_of_nat/
numoc-rec
nat-basic2
/
succ
nat-basic2
/
one-03
nat-basic1
/
neq-01
nat-basic2
/
ax-08
nat-basic1
/
f
nat
/
anti
nat
/
lels
nat-basic2
/
c
nat-basic2
/
a
nat-basic2
/
one
nat
/
lels-03
nat-basic1
/
lf-02
nat-basic1
/
ref-1
nat-basic1
/
ls-02