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Theorem List for New Foundations Explorer - 6201-6300   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremncpw1pwneg 6201 The cardinality of a unit power class is not equal to the cardinality of the power class. Theorem XI.2.4 of [Rosser] p. 372. (Contributed by SF, 10-Mar-2015.)
(A VNc 1ANc A)
 
Theoremltcpw1pwg 6202 The cardinality of a unit power class is strictly less than the cardinality of the power class. Theorem XI.2.17 of [Rosser] p. 376. (Contributed by SF, 10-Mar-2015.)
(A VNc 1A <c Nc A)
 
Theoremsbthlem1 6203 Lemma for sbth 6206. Set up similarity with a range. Theorem XI.1.14 of [Rosser] p. 350. (Contributed by SF, 11-Mar-2015.)
R V    &   X V    &   G = Clos1 ((X ran R), R)    &   A = (XG)    &   B = (X G)    &   C = (ran RG)    &   D = (ran R G)       (((Fun R Fun R) (X dom R ran R X)) → ran RX)
 
Theoremsbthlem2 6204 Lemma for sbth 6206. Eliminate hypotheses from sbthlem1 6203. Theorem XI.1.14 of [Rosser] p. 350. (Contributed by SF, 10-Mar-2015.)
R V       (((Fun R Fun R) (B V B dom R ran R B)) → ran RB)
 
Theoremsbthlem3 6205 Lemma for sbth 6206. If A is equinumerous with a subset of B and vice-versa, then A is equinumerous with B. Theorem XI.1.15 of [Rosser] p. 353. (Contributed by SF, 10-Mar-2015.)
(((AC C B) (BD D A)) → AB)
 
Theoremsbth 6206 The Schroder-Bernstein Theorem. This theorem gives the antisymmetry law for cardinal less than or equal. Translated out, it means that, if A is no larger than B and B is no larger than A, then Nc A = Nc B. Theorem XI.2.20 of [Rosser] p. 376. (Contributed by SF, 11-Mar-2015.)
((A NC B NC ) → ((Ac B Bc A) → A = B))
 
Theoremltlenlec 6207 Cardinal less than is equivalent to one-way cardinal less than or equal. Theorem XI.2.21 of [Rosser] p. 377. (Contributed by SF, 11-Mar-2015.)
((M NC N NC ) → (M <c N ↔ (Mc N ¬ Nc M)))
 
Theoremaddlec 6208 For non-empty sets, cardinal sum always increases cardinal less than or equal. Theorem XI.2.19 of [Rosser] p. 376. (Contributed by SF, 11-Mar-2015.)
((M V N W (M +c N) ≠ ) → Mc (M +c N))
 
Theoremaddlecncs 6209 For cardinals, cardinal sum always increases cardinal less than or equal. Corollary of theorem XI.2.19 of [Rosser] p. 376. (Contributed by SF, 11-Mar-2015.)
((M NC N NC ) → Mc (M +c N))
 
Theoremdflec2 6210* Cardinal less than or equal in terms of cardinal addition. Theorem XI.2.22 of [Rosser] p. 377. (Contributed by SF, 11-Mar-2015.)
((M NC N NC ) → (Mc Np NC N = (M +c p)))
 
Theoremlectr 6211 Cardinal less than or equal is transitive. (Contributed by SF, 12-Mar-2015.)
((A NC B NC C NC ) → ((Ac B Bc C) → Ac C))
 
Theoremleltctr 6212 Transitivity law for cardinal less than or equal and less than. (Contributed by SF, 16-Mar-2015.)
((A NC B NC C NC ) → ((Ac B B <c C) → A <c C))
 
Theoremlecponc 6213 Cardinal less than or equal partially orders the cardinals. (Contributed by SF, 12-Mar-2015.)
c Po NC
 
Theoremleaddc1 6214 Addition law for cardinal less than. Theorem XI.2.23 of [Rosser] p. 377. (Contributed by SF, 12-Mar-2015.)
(((M NC N NC P NC ) Mc N) → (M +c P) ≤c (N +c P))
 
Theoremleaddc2 6215 Addition law for cardinal less than. Theorem XI.2.23 of [Rosser] p. 377. (Contributed by SF, 12-Mar-2015.)
(((M NC N NC P NC ) Nc P) → (M +c N) ≤c (M +c P))
 
Theoremnc0le1 6216 Any cardinal is either zero or no greater than one. Theorem XI.2.24 of [Rosser] p. 377. (Contributed by SF, 12-Mar-2015.)
(N NC → (N = 0c 1cc N))
 
Theoremnc0suc 6217* Any cardinal is either zero or the successor of a cardinal. Corollary of theorem XI.2.24 of [Rosser] p. 377. (Contributed by SF, 12-Mar-2015.)
(N NC → (N = 0c m NC N = (m +c 1c)))
 
Theoremleconnnc 6218 Cardinal less than or equal is total over the naturals. (Contributed by SF, 12-Mar-2015.)
((A Nn B Nn ) → (Ac B Bc A))
 
Theoremaddceq0 6219 The sum of two cardinals is zero iff both addends are zero. (Contributed by SF, 12-Mar-2015.)
((A NC B NC ) → ((A +c B) = 0c ↔ (A = 0c B = 0c)))
 
Theoremce2lt 6220 Ordering law for cardinal exponentiation to two. Theorem XI.2.71 of [Rosser] p. 390. (Contributed by SF, 13-Mar-2015.)
((M NC (Mc 0c) NC ) → M <c (2cc M))
 
Theoremdflec3 6221* Another potential definition of cardinal inequality. (Contributed by SF, 23-Mar-2015.)
((M NC N NC ) → (Mc Na M b N f f:a1-1b))
 
Theoremnclenc 6222* Comparison rule for cardinalities. (Contributed by SF, 24-Mar-2015.)
A V    &   B V       ( Nc Ac Nc Bf f:A1-1B)
 
Theoremlenc 6223* Less than or equal condition for the cardinality of a number. (Contributed by SF, 18-Mar-2015.)
A V       (M NC → (Mc Nc Ax M x A))
 
Theoremtcnc 6224 Compute the T-raising of a cardinality. (Contributed by SF, 4-Mar-2015.)
A V        Tc Nc A = Nc 1A
 
Theoremtcncv 6225 Compute the T-raising of the cardinality of the universe. Part of Theorem 5.2 of [Specker] p. 973. (Contributed by SF, 4-Mar-2015.)
Tc Nc V = Nc 1c
 
Theoremtcnc1c 6226 Compute the T-raising of the cardinality of one. Part of Theorem 5.2 of [Specker] p. 973. (Contributed by SF, 4-Mar-2015.)
Tc Nc 1c = Nc 11c
 
Theoremtc11 6227 Cardinal T is one-to-one. Based on theorem 2.4 of [Specker] p. 972. (Contributed by SF, 10-Mar-2015.)
((M NC N NC ) → ( Tc M = Tc NM = N))
 
Theoremtaddc 6228* T raising rule for cardinal sum. (Contributed by SF, 11-Mar-2015.)
(((A NC B NC X NC ) Tc A = ( Tc B +c X)) → c NC X = Tc c)
 
Theoremtlecg 6229 T-raising perserves order for cardinals. Theorem 5.5 of [Specker] p. 973. (Contributed by SF, 11-Mar-2015.)
((M NC N NC ) → (Mc NTc Mc Tc N))
 
Theoremletc 6230* If a cardinal is less than or equal to a T-raising, then it is also a T-raising. Theorem 5.6 of [Specker] p. 973. (Contributed by SF, 11-Mar-2015.)
((M NC N NC Mc Tc N) → p NC M = Tc p)
 
Theoremce0t 6231* If (Mc 0c) is a cardinal, then M is a T-raising of some cardinal. (Contributed by SF, 17-Mar-2015.)
((M NC (Mc 0c) NC ) → n NC M = Tc n)
 
Theoremce2le 6232 Partial ordering law for base two cardinal exponentiation. Theorem 4.8 of [Specker] p. 973. (Contributed by SF, 16-Mar-2015.)
(((M NC N NC (Nc 0c) NC ) Mc N) → (2cc M) ≤c (2cc N))
 
Theoremcet 6233 The exponent of a T-raising to a T-raising is always a cardinal. (Contributed by SF, 13-Mar-2015.)
((M NC N NC ) → ( Tc Mc Tc N) NC )
 
Theoremce2t 6234 The exponent of two to a T-raising is always a cardinal. Theorem 5.8 of [Specker] p. 973. (Contributed by SF, 13-Mar-2015.)
(M NC → (2cc Tc M) NC )
 
Theoremtce2 6235 Distributive law for T-raising and cardinal exponentiation to two. (Contributed by SF, 13-Mar-2015.)
((M NC (Mc 0c) NC ) → Tc (2cc M) = (2cc Tc M))
 
Theoremte0c 6236 A T-raising raised to zero is always a cardinal. (Contributed by SF, 16-Mar-2015.)
(M NC → ( Tc Mc 0c) NC )
 
Theoremce0tb 6237* (Mc 0c) is a cardinal iff M is a T-raising of some cardinal. (Contributed by SF, 17-Mar-2015.)
(M NC → ((Mc 0c) NCn NC M = Tc n))
 
Theoremce0lenc1 6238 Cardinal exponentiation to zero is a cardinal iff the number is less than the size of cardinal one. (Contributed by SF, 18-Mar-2015.)
(M NC → ((Mc 0c) NCMc Nc 1c))
 
Theoremtlenc1c 6239 A T-raising is less than or equal to the cardinality of cardinal one. (Contributed by SF, 16-Mar-2015.)
(M NCTc Mc Nc 1c)
 
Theorem1ne0c 6240 Cardinal one is not zero. (Contributed by SF, 4-Mar-2015.)
1c ≠ 0c
 
Theorem2ne0c 6241 Cardinal two is not zero. (Contributed by SF, 4-Mar-2015.)
2c ≠ 0c
 
Theoremfinnc 6242 A set is finite iff its cardinality is a natural. (Contributed by SF, 18-Mar-2015.)
(A FinNc A Nn )
 
Theoremtcfnex 6243 The stratified T raising function is a set. (Contributed by SF, 18-Mar-2015.)
TcFn V
 
Theoremfntcfn 6244 Functionhood statement for the stratified T-raising function. (Contributed by SF, 18-Mar-2015.)
TcFn Fn 1c
 
Theorembrtcfn 6245 Binary relationship form of the stratified T-raising function. (Contributed by SF, 18-Mar-2015.)
A V       ({A}TcFnBB = Tc A)
 
Theoremncfin 6246 The cardinality of a set is a natural iff the set is finite. (Contributed by SF, 19-Mar-2015.)
A V       ( Nc A NnA Fin )
 
Theoremnclennlem1 6247* Lemma for nclenn 6248. Set up stratification for induction. (Contributed by SF, 19-Mar-2015.)
{x n NC (nc xn Nn )} V
 
Theoremnclenn 6248 A cardinal that is less than or equal to a natural is a natural. Theorem XI.3.3 of [Rosser] p. 391. (Contributed by SF, 19-Mar-2015.)
((M NC N Nn Mc N) → M Nn )
 
Theoremaddcdi 6249 Distributivity law for cardinal addition and multiplication. Theorem XI.2.31 of [Rosser] p. 379. (Contributed by Scott Fenton, 31-Jul-2019.)
((A NC B NC C NC ) → (A ·c (B +c C)) = ((A ·c B) +c (A ·c C)))
 
Theoremaddcdir 6250 Distributivity law for cardinal addition and multiplication. Theorem XI.2.30 of [Rosser] p. 379. (Contributed by Scott Fenton, 31-Jul-2019.)
((A NC B NC C NC ) → ((A +c B) ·c C) = ((A ·c C) +c (B ·c C)))
 
Theoremmuc0or 6251 The cardinal product of two cardinal numbers is zero iff one of the numbers is zero. Biconditional form of theorem XI.2.34 of [Rosser] p. 380. (Contributed by Scott Fenton, 31-Jul-2019.)
((A NC B NC ) → ((A ·c B) = 0c ↔ (A = 0c B = 0c)))
 
Theoremlemuc1 6252 Multiplication law for cardinal less than. Theorem XI.2.35 of [Rosser] p. 380. (Contributed by Scott Fenton, 31-Jul-2019.)
(((A NC B NC C NC ) Ac B) → (A ·c C) ≤c (B ·c C))
 
Theoremlemuc2 6253 Multiplication law for cardinal less than. (Contributed by Scott Fenton, 31-Jul-2019.)
(((A NC B NC C NC ) Bc C) → (A ·c B) ≤c (A ·c C))
 
Theoremncslemuc 6254 A cardinal is less than or equal to its product with another. Theorem XI.2.36 of [Rosser] p. 381. (Contributed by Scott Fenton, 31-Jul-2019.)
((M NC N NC N ≠ 0c) → Mc (M ·c N))
 
Theoremncvsq 6255 The product of the cardinality of V squared is just the cardinality of V. Theorem XI.2.37 of [Rosser] p. 381. (Contributed by Scott Fenton, 31-Jul-2019.)
( Nc V ·c Nc V) = Nc V
 
Theoremvvsqenvv 6256 There are exactly as many ordered pairs as there are sets. Corollary to theorem XI.2.37 of [Rosser] p. 381. (Contributed by Scott Fenton, 31-Jul-2019.)
(V × V) ≈ V
 
Theorem0lt1c 6257 Cardinal one is strictly greater than cardinal zero. (Contributed by Scott Fenton, 1-Aug-2019.)
0c <c 1c
 
Theoremcsucex 6258 The function mapping x to its cardinal successor exists. (Contributed by Scott Fenton, 30-Jul-2019.)
(x V (x +c 1c)) V
 
Theorembrcsuc 6259* Binary relationship form of the successor mapping function. (Contributed by Scott Fenton, 2-Aug-2019.)
A V    &   B V       (A(x V (x +c 1c))BB = (A +c 1c))
 
Theoremnnltp1clem1 6260 Lemma for nnltp1c 6261. Set up stratification. (Contributed by SF, 25-Mar-2015.)
{x x <c (x +c 1c)} V
 
Theoremnnltp1c 6261 Any natural is less than one plus itself. (Contributed by SF, 25-Mar-2015.)
(N NnN <c (N +c 1c))
 
Theoremaddccan2nclem1 6262* Lemma for addccan2nc 6264. Stratification helper theorem. (Contributed by Scott Fenton, 2-Aug-2019.)
(x( AddC (1st (V × {n})))yy = (x +c n))
 
Theoremaddccan2nclem2 6263* Lemma for addccan2nc 6264. Establish stratification for induction. (Contributed by Scott Fenton, 2-Aug-2019.)
((N V P W) → {x ((x +c N) = (x +c P) → N = P)} V)
 
Theoremaddccan2nc 6264 Cancellation law for addition over the cardinal numbers. Biconditional form of theorem XI.3.2 of [Rosser] p. 391. (Contributed by Scott Fenton, 2-Aug-2019.)
((M Nn N NC P NC ) → ((M +c N) = (M +c P) ↔ N = P))
 
Theoremlecadd2 6265 Cardinal addition preserves cardinal less than. Biconditional form of corollary 4 of theorem XI.3.2 of [Rosser] p. 391. (Contributed by Scott Fenton, 2-Aug-2019.)
((M Nn N NC P NC ) → ((M +c N) ≤c (M +c P) ↔ Nc P))
 
Theoremncslesuc 6266 Relationship between successor and cardinal less than or equal. (Contributed by Scott Fenton, 3-Aug-2019.)
((M NC N NC ) → (Mc (N +c 1c) ↔ (Mc N M = (N +c 1c))))
 
Theoremnmembers1lem1 6267* Lemma for nmembers1 6270. Set up stratification. (Contributed by SF, 25-Mar-2015.)
{x {m Nn (1cc m mc x)} Tc Tc x} V
 
Theoremnmembers1lem2 6268 Lemma for nmembers1 6270. The set of all elements between one and zero is empty. (Contributed by Scott Fenton, 1-Aug-2019.)
{m Nn (1cc m mc 0c)} 0c
 
Theoremnmembers1lem3 6269* Lemma for nmembers1 6270. If the interval from one to a natural is in a given natural, extending it by one puts it in the next natural. (Contributed by Scott Fenton, 3-Aug-2019.)
((A Nn B Nn ) → ({m Nn (1cc m mc A)} B → {m Nn (1cc m mc (A +c 1c))} (B +c 1c)))
 
Theoremnmembers1 6270* Count the number of elements in a natural interval. From nmembers1lem2 6268 and nmembers1lem3 6269, we would expect to arrive at {m Nn (1cc m mc N)} N, but this proposition is not stratifiable. Instead, we arrive at the weaker conclusion below. We can arrive at the earlier proposition once we add the Axiom of Counting, which we will do later. (Contributed by Scott Fenton, 3-Aug-2019.)
(N Nn → {m Nn (1cc m mc N)} Tc Tc N)
 
2.4.6  Specker's disproof of the axiom of choice
 
Syntaxcspac 6271 Extend the definition of a class to include the special set generator for the axiom of choice.
class Spac
 
Definitiondf-spac 6272* Define the special class generator for the disproof of the axiom of choice. Definition 6.1 of [Specker] p. 973. (Contributed by SF, 3-Mar-2015.)
Spac = (m NC Clos1 ({m}, {x, y (x NC y NC y = (2cc x))}))
 
Theoremnncdiv3lem1 6273 Lemma for nncdiv3 6275. Set up a helper for stratification. (Contributed by SF, 3-Mar-2015.)
(n, b ran ( Ins3 ((ran (1st ⊗ (1st ∩ 2nd )) ⊗ 2nd ) “ AddC ) ∩ (((1st 1st ) ⊗ ((2nd 1st ) ⊗ 2nd )) “ AddC )) ↔ b = ((n +c n) +c n))
 
Theoremnncdiv3lem2 6274* Lemma for nncdiv3 6275. Set up stratification for induction. (Contributed by SF, 2-Mar-2015.)
{a n Nn (a = ((n +c n) +c n) a = (((n +c n) +c n) +c 1c) a = (((n +c n) +c n) +c 2c))} V
 
Theoremnncdiv3 6275* Divisibility by three rule for finite cardinals. Part of Theorem 3.4 of [Specker] p. 973. (Contributed by SF, 2-Mar-2015.)
(A Nnn Nn (A = ((n +c n) +c n) A = (((n +c n) +c n) +c 1c) A = (((n +c n) +c n) +c 2c)))
 
Theoremnnc3n3p1 6276 Three times a natural is not one more than three times a natural. Another part of Theorem 3.4 of [Specker] p. 973. (Contributed by SF, 13-Mar-2015.)
((A Nn B Nn ) → ¬ ((A +c A) +c A) = (((B +c B) +c B) +c 1c))
 
Theoremnnc3n3p2 6277 Three times a natural is not two more than three times a natural. Another part of Theorem 3.4 of [Specker] p. 973. (Contributed by SF, 12-Mar-2015.)
((A Nn B Nn ) → ¬ ((A +c A) +c A) = (((B +c B) +c B) +c 2c))
 
Theoremnnc3p1n3p2 6278 One more than three times a natural is not two more than three times a natural. Final part of Theorem 3.4 of [Specker] p. 973. (Contributed by SF, 12-Mar-2015.)
((A Nn B Nn ) → ¬ (((A +c A) +c A) +c 1c) = (((B +c B) +c B) +c 2c))
 
Theoremspacvallem1 6279* Lemma for spacval 6280. Set up stratification for the recursive relationship. (Contributed by SF, 6-Mar-2015.)
{x, y (x NC y NC y = (2cc x))} V
 
Theoremspacval 6280* The value of the special set generator. (Contributed by SF, 4-Mar-2015.)
(N NC → ( SpacN) = Clos1 ({N}, {x, y (x NC y NC y = (2cc x))}))
 
Theoremfnspac 6281 The special set generator is a function over the cardinals. (Contributed by SF, 18-Mar-2015.)
Spac Fn NC
 
Theoremspacssnc 6282 The special set generator generates a set of cardinals. (Contributed by SF, 13-Mar-2015.)
(N NC → ( SpacN) NC )
 
Theoremspacid 6283 The initial value of the special set generator is an element. (Contributed by SF, 13-Mar-2015.)
(M NCM ( SpacM))
 
Theoremspaccl 6284 Closure law for the special set generator. (Contributed by SF, 13-Mar-2015.)
((M NC N ( SpacM) (Nc 0c) NC ) → (2cc N) ( SpacM))
 
Theoremspacind 6285* Inductive law for the special set generator. (Contributed by SF, 13-Mar-2015.)
(((M NC S V) (M S x ( SpacM)((x S (xc 0c) NC ) → (2cc x) S))) → ( SpacM) S)
 
Theoremspacis 6286* Induction scheme for the special set generator. (Contributed by SF, 13-Mar-2015.)
{x φ} V    &   (x = M → (φψ))    &   (x = y → (φχ))    &   (x = (2cc y) → (φθ))    &   (x = N → (φτ))    &   (M NCψ)    &   (((M NC y ( SpacM)) ((yc 0c) NC χ)) → θ)       ((M NC N ( SpacM)) → τ)
 
Theoremnchoicelem1 6287 Lemma for nchoice 6306. A finite cardinal is not one more than its T-raising. (Contributed by SF, 3-Mar-2015.)
(A Nn → ¬ A = ( Tc A +c 1c))
 
Theoremnchoicelem2 6288 Lemma for nchoice 6306. A finite cardinal is not two more than its T-raising. (Contributed by SF, 12-Mar-2015.)
(A Nn → ¬ A = ( Tc A +c 2c))
 
Theoremnchoicelem3 6289 Lemma for nchoice 6306. Compute the value of Spac when the argument is not exponentiable. Theorem 6.2 of [Specker] p. 973. (Contributed by SF, 13-Mar-2015.)
((M NC ¬ (Mc 0c) NC ) → ( SpacM) = {M})
 
Theoremnchoicelem4 6290 Lemma for nchoice 6306. The initial value of Spac is a minimum value. Theorem 6.4 of [Specker] p. 973. (Contributed by SF, 13-Mar-2015.)
((M NC N ( SpacM)) → Mc N)
 
Theoremnchoicelem5 6291 Lemma for nchoice 6306. A cardinal is not a member of the special set of itself raised to two. Theorem 6.5 of [Specker] p. 973. (Contributed by SF, 13-Mar-2015.)
((M NC (Mc 0c) NC ) → ¬ M ( Spac ‘(2cc M)))
 
Theoremnchoicelem6 6292 Lemma for nchoice 6306. Split the special set generator into base and inductive values. Theorem 6.6 of [Specker] p. 973. (Contributed by SF, 13-Mar-2015.)
((M NC (Mc 0c) NC ) → ( SpacM) = ({M} ∪ ( Spac ‘(2cc M))))
 
Theoremnchoicelem7 6293 Lemma for nchoice 6306. Calculate the cardinality of a special set generator. Theorem 6.7 of [Specker] p. 974. (Contributed by SF, 13-Mar-2015.)
((M NC (Mc 0c) NC ) → Nc ( SpacM) = ( Nc ( Spac ‘(2cc M)) +c 1c))
 
Theoremnchoicelem8 6294 Lemma for nchoice 6306. An anti-closure condition for cardinal exponentiation to zero. Theorem 4.5 of [Specker] p. 973. (Contributed by SF, 18-Mar-2015.)
(( ≤c We NC M NC ) → (¬ (Mc 0c) NCNc 1c <c M))
 
Theoremnchoicelem9 6295 Lemma for nchoice 6306. Calculate the cardinality of the special set generator when near the end of raisability. Theorem 6.8 of [Specker] p. 974. (Contributed by SF, 18-Mar-2015.)
(( ≤c We NC M NC ¬ (Mc 0c) NC ) → ( Nc ( SpacTc M) = 2c Nc ( SpacTc M) = 3c))
 
Theoremnchoicelem10 6296 Lemma for nchoice 6306. Stratification helper lemma. (Contributed by SF, 18-Mar-2015.)
S V    &   X V       (c, X ∼ (( Ins3 S Ins2 ∼ ran ( S ⊗ ( S Fix ( S ImageS)))) “ 1c) ↔ c = Clos1 (X, S))
 
Theoremnchoicelem11 6297* Lemma for nchoice 6306. Set up stratification for nchoicelem12 6298. (Contributed by SF, 18-Mar-2015.)
{t m NC (t = Nc ( SpacTc m) → Nc ( Spacm) Nn )} V
 
Theoremnchoicelem12 6298 Lemma for nchoice 6306. If the T-raising of a cardinal yields a finite special set, then so does the initial set. Theorem 7.1 of [Specker] p. 974. (Contributed by SF, 18-Mar-2015.)
((M NC ( SpacTc M) Fin ) → ( SpacM) Fin )
 
Theoremnchoicelem13 6299 Lemma for nchoice 6306. The cardinality of a special set is at least one. (Contributed by SF, 18-Mar-2015.)
(M NC → 1cc Nc ( SpacM))
 
Theoremnchoicelem14 6300 Lemma for nchoice 6306. When the special set generator yields a singleton, then the cardinal is not raisable. (Contributed by SF, 19-Mar-2015.)
((M NC Nc ( SpacM) = 1c) → ¬ (Mc 0c) NC )
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