Theorem List for Intuitionistic Logic Explorer - 6901-7000 *Has distinct variable
group(s)
| Type | Label | Description |
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| Theorem | fndmeng 6901 |
A function is equinumerate to its domain. (Contributed by Paul Chapman,
22-Jun-2011.)
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| Theorem | mapsnen 6902 |
Set exponentiation to a singleton exponent is equinumerous to its base.
Exercise 4.43 of [Mendelson] p. 255.
(Contributed by NM, 17-Dec-2003.)
(Revised by Mario Carneiro, 15-Nov-2014.)
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| Theorem | map1 6903 |
Set exponentiation: ordinal 1 to any set is equinumerous to ordinal 1.
Exercise 4.42(b) of [Mendelson] p.
255. (Contributed by NM,
17-Dec-2003.)
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| Theorem | en2sn 6904 |
Two singletons are equinumerous. (Contributed by NM, 9-Nov-2003.)
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| Theorem | snfig 6905 |
A singleton is finite. For the proper class case, see snprc 3697.
(Contributed by Jim Kingdon, 13-Apr-2020.)
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| Theorem | fiprc 6906 |
The class of finite sets is a proper class. (Contributed by Jeff
Hankins, 3-Oct-2008.)
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| Theorem | unen 6907 |
Equinumerosity of union of disjoint sets. Theorem 4 of [Suppes] p. 92.
(Contributed by NM, 11-Jun-1998.) (Revised by Mario Carneiro,
26-Apr-2015.)
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| Theorem | en2prd 6908 |
Two proper unordered pairs are equinumerous. (Contributed by
BTernaryTau, 23-Dec-2024.)
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| Theorem | rex2dom 6909* |
A set that has at least 2 different members dominates ordinal 2.
(Contributed by BTernaryTau, 30-Dec-2024.)
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| Theorem | enpr2d 6910 |
A pair with distinct elements is equinumerous to ordinal two.
(Contributed by Rohan Ridenour, 3-Aug-2023.)
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| Theorem | en2 6911* |
A set equinumerous to ordinal 2 is an unordered pair. (Contributed by
Mario Carneiro, 5-Jan-2016.)
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| Theorem | ssct 6912 |
A subset of a set dominated by is dominated by .
(Contributed by Thierry Arnoux, 31-Jan-2017.)
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| Theorem | 1domsn 6913 |
A singleton (whether of a set or a proper class) is dominated by one.
(Contributed by Jim Kingdon, 1-Mar-2022.)
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| Theorem | enm 6914* |
A set equinumerous to an inhabited set is inhabited. (Contributed by
Jim Kingdon, 19-May-2020.)
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| Theorem | xpsnen 6915 |
A set is equinumerous to its Cartesian product with a singleton.
Proposition 4.22(c) of [Mendelson] p.
254. (Contributed by NM,
4-Jan-2004.) (Revised by Mario Carneiro, 15-Nov-2014.)
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| Theorem | xpsneng 6916 |
A set is equinumerous to its Cartesian product with a singleton.
Proposition 4.22(c) of [Mendelson] p.
254. (Contributed by NM,
22-Oct-2004.)
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| Theorem | xp1en 6917 |
One times a cardinal number. (Contributed by NM, 27-Sep-2004.) (Revised
by Mario Carneiro, 29-Apr-2015.)
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| Theorem | endisj 6918* |
Any two sets are equinumerous to disjoint sets. Exercise 4.39 of
[Mendelson] p. 255. (Contributed by
NM, 16-Apr-2004.)
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| Theorem | xpcomf1o 6919* |
The canonical bijection from   to   .
(Contributed by Mario Carneiro, 23-Apr-2014.)
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| Theorem | xpcomco 6920* |
Composition with the bijection of xpcomf1o 6919 swaps the arguments to a
mapping. (Contributed by Mario Carneiro, 30-May-2015.)
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| Theorem | xpcomen 6921 |
Commutative law for equinumerosity of Cartesian product. Proposition
4.22(d) of [Mendelson] p. 254.
(Contributed by NM, 5-Jan-2004.)
(Revised by Mario Carneiro, 15-Nov-2014.)
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| Theorem | xpcomeng 6922 |
Commutative law for equinumerosity of Cartesian product. Proposition
4.22(d) of [Mendelson] p. 254.
(Contributed by NM, 27-Mar-2006.)
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| Theorem | xpsnen2g 6923 |
A set is equinumerous to its Cartesian product with a singleton on the
left. (Contributed by Stefan O'Rear, 21-Nov-2014.)
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| Theorem | xpassen 6924 |
Associative law for equinumerosity of Cartesian product. Proposition
4.22(e) of [Mendelson] p. 254.
(Contributed by NM, 22-Jan-2004.)
(Revised by Mario Carneiro, 15-Nov-2014.)
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| Theorem | xpdom2 6925 |
Dominance law for Cartesian product. Proposition 10.33(2) of
[TakeutiZaring] p. 92.
(Contributed by NM, 24-Jul-2004.) (Revised by
Mario Carneiro, 15-Nov-2014.)
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| Theorem | xpdom2g 6926 |
Dominance law for Cartesian product. Theorem 6L(c) of [Enderton]
p. 149. (Contributed by Mario Carneiro, 26-Apr-2015.)
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| Theorem | xpdom1g 6927 |
Dominance law for Cartesian product. Theorem 6L(c) of [Enderton]
p. 149. (Contributed by NM, 25-Mar-2006.) (Revised by Mario Carneiro,
26-Apr-2015.)
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| Theorem | xpdom3m 6928* |
A set is dominated by its Cartesian product with an inhabited set.
Exercise 6 of [Suppes] p. 98.
(Contributed by Jim Kingdon,
15-Apr-2020.)
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| Theorem | xpdom1 6929 |
Dominance law for Cartesian product. Theorem 6L(c) of [Enderton]
p. 149. (Contributed by NM, 28-Sep-2004.) (Revised by NM,
29-Mar-2006.) (Revised by Mario Carneiro, 7-May-2015.)
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| Theorem | pw2f1odclem 6930* |
Lemma for pw2f1odc 6931. (Contributed by Mario Carneiro,
6-Oct-2014.)
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DECID                   
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| Theorem | pw2f1odc 6931* |
The power set of a set is equinumerous to set exponentiation with an
unordered pair base of ordinal 2. Generalized from Proposition 10.44 of
[TakeutiZaring] p. 96.
(Contributed by Mario Carneiro, 6-Oct-2014.)
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DECID    
                
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| Theorem | fopwdom 6932 |
Covering implies injection on power sets. (Contributed by Stefan
O'Rear, 6-Nov-2014.) (Revised by Mario Carneiro, 24-Jun-2015.)
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| Theorem | 0domg 6933 |
Any set dominates the empty set. (Contributed by NM, 26-Oct-2003.)
(Revised by Mario Carneiro, 26-Apr-2015.)
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| Theorem | dom0 6934 |
A set dominated by the empty set is empty. (Contributed by NM,
22-Nov-2004.)
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| Theorem | 0dom 6935 |
Any set dominates the empty set. (Contributed by NM, 26-Oct-2003.)
(Revised by Mario Carneiro, 26-Apr-2015.)
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| Theorem | enen1 6936 |
Equality-like theorem for equinumerosity. (Contributed by NM,
18-Dec-2003.)
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| Theorem | enen2 6937 |
Equality-like theorem for equinumerosity. (Contributed by NM,
18-Dec-2003.)
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| Theorem | domen1 6938 |
Equality-like theorem for equinumerosity and dominance. (Contributed by
NM, 8-Nov-2003.)
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| Theorem | domen2 6939 |
Equality-like theorem for equinumerosity and dominance. (Contributed by
NM, 8-Nov-2003.)
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| 2.6.29 Equinumerosity (cont.)
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| Theorem | xpf1o 6940* |
Construct a bijection on a Cartesian product given bijections on the
factors. (Contributed by Mario Carneiro, 30-May-2015.)
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| Theorem | xpen 6941 |
Equinumerosity law for Cartesian product. Proposition 4.22(b) of
[Mendelson] p. 254. (Contributed by
NM, 24-Jul-2004.)
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| Theorem | mapen 6942 |
Two set exponentiations are equinumerous when their bases and exponents
are equinumerous. Theorem 6H(c) of [Enderton] p. 139. (Contributed by
NM, 16-Dec-2003.) (Proof shortened by Mario Carneiro, 26-Apr-2015.)
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| Theorem | mapdom1g 6943 |
Order-preserving property of set exponentiation. (Contributed by Jim
Kingdon, 15-Jul-2022.)
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| Theorem | mapxpen 6944 |
Equinumerosity law for double set exponentiation. Proposition 10.45 of
[TakeutiZaring] p. 96.
(Contributed by NM, 21-Feb-2004.) (Revised by
Mario Carneiro, 24-Jun-2015.)
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| Theorem | xpmapenlem 6945* |
Lemma for xpmapen 6946. (Contributed by NM, 1-May-2004.) (Revised
by
Mario Carneiro, 16-Nov-2014.)
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| Theorem | xpmapen 6946 |
Equinumerosity law for set exponentiation of a Cartesian product.
Exercise 4.47 of [Mendelson] p. 255.
(Contributed by NM, 23-Feb-2004.)
(Proof shortened by Mario Carneiro, 16-Nov-2014.)
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| Theorem | ssenen 6947* |
Equinumerosity of equinumerous subsets of a set. (Contributed by NM,
30-Sep-2004.) (Revised by Mario Carneiro, 16-Nov-2014.)
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| 2.6.30 Pigeonhole Principle
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| Theorem | phplem1 6948 |
Lemma for Pigeonhole Principle. If we join a natural number to itself
minus an element, we end up with its successor minus the same element.
(Contributed by NM, 25-May-1998.)
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| Theorem | phplem2 6949 |
Lemma for Pigeonhole Principle. A natural number is equinumerous to its
successor minus one of its elements. (Contributed by NM, 11-Jun-1998.)
(Revised by Mario Carneiro, 16-Nov-2014.)
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| Theorem | phplem3 6950 |
Lemma for Pigeonhole Principle. A natural number is equinumerous to its
successor minus any element of the successor. For a version without the
redundant hypotheses, see phplem3g 6952. (Contributed by NM,
26-May-1998.)
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| Theorem | phplem4 6951 |
Lemma for Pigeonhole Principle. Equinumerosity of successors implies
equinumerosity of the original natural numbers. (Contributed by NM,
28-May-1998.) (Revised by Mario Carneiro, 24-Jun-2015.)
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| Theorem | phplem3g 6952 |
A natural number is equinumerous to its successor minus any element of
the successor. Version of phplem3 6950 with unnecessary hypotheses
removed. (Contributed by Jim Kingdon, 1-Sep-2021.)
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| Theorem | nneneq 6953 |
Two equinumerous natural numbers are equal. Proposition 10.20 of
[TakeutiZaring] p. 90 and its
converse. Also compare Corollary 6E of
[Enderton] p. 136. (Contributed by NM,
28-May-1998.)
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| Theorem | php5 6954 |
A natural number is not equinumerous to its successor. Corollary
10.21(1) of [TakeutiZaring] p. 90.
(Contributed by NM, 26-Jul-2004.)
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| Theorem | snnen2og 6955 |
A singleton   is never equinumerous with the ordinal
number 2. If
is a proper
class, see snnen2oprc 6956. (Contributed by Jim Kingdon,
1-Sep-2021.)
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| Theorem | snnen2oprc 6956 |
A singleton   is never equinumerous with the ordinal
number 2. If
is a set, see snnen2og 6955. (Contributed by Jim Kingdon,
1-Sep-2021.)
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| Theorem | 1nen2 6957 |
One and two are not equinumerous. (Contributed by Jim Kingdon,
25-Jan-2022.)
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| Theorem | phplem4dom 6958 |
Dominance of successors implies dominance of the original natural
numbers. (Contributed by Jim Kingdon, 1-Sep-2021.)
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| Theorem | php5dom 6959 |
A natural number does not dominate its successor. (Contributed by Jim
Kingdon, 1-Sep-2021.)
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| Theorem | nndomo 6960 |
Cardinal ordering agrees with natural number ordering. Example 3 of
[Enderton] p. 146. (Contributed by NM,
17-Jun-1998.)
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| Theorem | phpm 6961* |
Pigeonhole Principle. A natural number is not equinumerous to a proper
subset of itself. By "proper subset" here we mean that there
is an
element which is in the natural number and not in the subset, or in
symbols     (which is stronger than not being equal
in the absence of excluded middle). Theorem (Pigeonhole Principle) of
[Enderton] p. 134. The theorem is
so-called because you can't put n +
1 pigeons into n holes (if each hole holds only one pigeon). The
proof consists of lemmas phplem1 6948 through phplem4 6951, nneneq 6953, and
this final piece of the proof. (Contributed by NM, 29-May-1998.)
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| Theorem | phpelm 6962 |
Pigeonhole Principle. A natural number is not equinumerous to an
element of itself. (Contributed by Jim Kingdon, 6-Sep-2021.)
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| Theorem | phplem4on 6963 |
Equinumerosity of successors of an ordinal and a natural number implies
equinumerosity of the originals. (Contributed by Jim Kingdon,
5-Sep-2021.)
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| 2.6.31 Finite sets
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| Theorem | fict 6964 |
A finite set is dominated by . Also see finct 7217. (Contributed
by Thierry Arnoux, 27-Mar-2018.)
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| Theorem | fidceq 6965 |
Equality of members of a finite set is decidable. This may be
counterintuitive: cannot any two sets be elements of a finite set?
Well, to show, for example, that    is finite would require
showing it is equinumerous to or to but to show that you'd
need to know
or , respectively.
(Contributed by
Jim Kingdon, 5-Sep-2021.)
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| Theorem | fidifsnen 6966 |
All decrements of a finite set are equinumerous. (Contributed by Jim
Kingdon, 9-Sep-2021.)
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| Theorem | fidifsnid 6967 |
If we remove a single element from a finite set then put it back in, we
end up with the original finite set. This strengthens difsnss 3778 from
subset to equality when the set is finite. (Contributed by Jim Kingdon,
9-Sep-2021.)
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| Theorem | nnfi 6968 |
Natural numbers are finite sets. (Contributed by Stefan O'Rear,
21-Mar-2015.)
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| Theorem | enfi 6969 |
Equinumerous sets have the same finiteness. (Contributed by NM,
22-Aug-2008.)
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| Theorem | enfii 6970 |
A set equinumerous to a finite set is finite. (Contributed by Mario
Carneiro, 12-Mar-2015.)
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| Theorem | ssfilem 6971* |
Lemma for ssfiexmid 6972. (Contributed by Jim Kingdon, 3-Feb-2022.)
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| Theorem | ssfiexmid 6972* |
If any subset of a finite set is finite, excluded middle follows. One
direction of Theorem 2.1 of [Bauer], p.
485. (Contributed by Jim
Kingdon, 19-May-2020.)
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| Theorem | infiexmid 6973* |
If the intersection of any finite set and any other set is finite,
excluded middle follows. (Contributed by Jim Kingdon, 5-Feb-2022.)
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| Theorem | domfiexmid 6974* |
If any set dominated by a finite set is finite, excluded middle follows.
(Contributed by Jim Kingdon, 3-Feb-2022.)
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| Theorem | dif1en 6975 |
If a set is
equinumerous to the successor of a natural number
, then with an element removed is
equinumerous to .
(Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Stefan O'Rear,
16-Aug-2015.)
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| Theorem | dif1enen 6976 |
Subtracting one element from each of two equinumerous finite sets.
(Contributed by Jim Kingdon, 5-Jun-2022.)
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| Theorem | fiunsnnn 6977 |
Adding one element to a finite set which is equinumerous to a natural
number. (Contributed by Jim Kingdon, 13-Sep-2021.)
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| Theorem | php5fin 6978 |
A finite set is not equinumerous to a set which adds one element.
(Contributed by Jim Kingdon, 13-Sep-2021.)
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| Theorem | fisbth 6979 |
Schroeder-Bernstein Theorem for finite sets. (Contributed by Jim
Kingdon, 12-Sep-2021.)
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| Theorem | 0fin 6980 |
The empty set is finite. (Contributed by FL, 14-Jul-2008.)
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| Theorem | fin0 6981* |
A nonempty finite set has at least one element. (Contributed by Jim
Kingdon, 10-Sep-2021.)
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| Theorem | fin0or 6982* |
A finite set is either empty or inhabited. (Contributed by Jim Kingdon,
30-Sep-2021.)
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| Theorem | diffitest 6983* |
If subtracting any set from a finite set gives a finite set, any
proposition of the form is
decidable. This is not a proof of
full excluded middle, but it is close enough to show we won't be able to
prove   . (Contributed by Jim
Kingdon,
8-Sep-2021.)
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| Theorem | findcard 6984* |
Schema for induction on the cardinality of a finite set. The inductive
hypothesis is that the result is true on the given set with any one
element removed. The result is then proven to be true for all finite
sets. (Contributed by Jeff Madsen, 2-Sep-2009.)
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| Theorem | findcard2 6985* |
Schema for induction on the cardinality of a finite set. The inductive
step shows that the result is true if one more element is added to the
set. The result is then proven to be true for all finite sets.
(Contributed by Jeff Madsen, 8-Jul-2010.)
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| Theorem | findcard2s 6986* |
Variation of findcard2 6985 requiring that the element added in the
induction step not be a member of the original set. (Contributed by
Paul Chapman, 30-Nov-2012.)
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| Theorem | findcard2d 6987* |
Deduction version of findcard2 6985. If you also need
(which
doesn't come for free due to ssfiexmid 6972), use findcard2sd 6988 instead.
(Contributed by SO, 16-Jul-2018.)
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| Theorem | findcard2sd 6988* |
Deduction form of finite set induction . (Contributed by Jim Kingdon,
14-Sep-2021.)
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| Theorem | diffisn 6989 |
Subtracting a singleton from a finite set produces a finite set.
(Contributed by Jim Kingdon, 11-Sep-2021.)
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| Theorem | diffifi 6990 |
Subtracting one finite set from another produces a finite set.
(Contributed by Jim Kingdon, 8-Sep-2021.)
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| Theorem | infnfi 6991 |
An infinite set is not finite. (Contributed by Jim Kingdon,
20-Feb-2022.)
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| Theorem | ominf 6992 |
The set of natural numbers is not finite. Although we supply this theorem
because we can, the more natural way to express " is infinite" is
which is an instance
of domrefg 6857. (Contributed by NM,
2-Jun-1998.)
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| Theorem | isinfinf 6993* |
An infinite set contains subsets of arbitrarily large finite
cardinality. (Contributed by Jim Kingdon, 15-Jun-2022.)
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| Theorem | ac6sfi 6994* |
Existence of a choice function for finite sets. (Contributed by Jeff
Hankins, 26-Jun-2009.) (Proof shortened by Mario Carneiro,
29-Jan-2014.)
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| Theorem | tridc 6995* |
A trichotomous order is decidable. (Contributed by Jim Kingdon,
5-Sep-2022.)
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DECID     |
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| Theorem | fimax2gtrilemstep 6996* |
Lemma for fimax2gtri 6997. The induction step. (Contributed by Jim
Kingdon, 5-Sep-2022.)
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| Theorem | fimax2gtri 6997* |
A finite set has a maximum under a trichotomous order. (Contributed
by Jim Kingdon, 5-Sep-2022.)
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| Theorem | finexdc 6998* |
Decidability of existence, over a finite set and defined by a decidable
proposition. (Contributed by Jim Kingdon, 12-Jul-2022.)
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   DECID  DECID    |
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| Theorem | dfrex2fin 6999* |
Relationship between universal and existential quantifiers over a finite
set. Remark in Section 2.2.1 of [Pierik], p. 8. Although Pierik does
not mention the decidability condition explicitly, it does say
"only
finitely many x to check" which means there must be some way of
checking
each value of x. (Contributed by Jim Kingdon, 11-Jul-2022.)
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   DECID   
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| Theorem | infm 7000* |
An infinite set is inhabited. (Contributed by Jim Kingdon,
18-Feb-2022.)
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