HomeHome Metamath Proof Explorer < Previous   Next >
Browser slow? Try the
Unicode version.

Jump to page: Contents + 1 1-100 2 101-200 3 201-300 4 301-400 5 401-500 6 501-600 7 601-700 8 701-800 9 801-900 10 901-1000 11 1001-1100 12 1101-1200 13 1201-1300 14 1301-1400 15 1401-1500 16 1501-1600 17 1601-1700 18 1701-1800 19 1801-1900 20 1901-2000 21 2001-2100 22 2101-2200 23 2201-2300 24 2301-2400 25 2401-2500 26 2501-2600 27 2601-2700 28 2701-2800 29 2801-2900 30 2901-3000 31 3001-3100 32 3101-3200 33 3201-3300 34 3301-3400 35 3401-3500 36 3501-3600 37 3601-3700 38 3701-3800 39 3801-3900 40 3901-4000 41 4001-4100 42 4101-4200 43 4201-4300 44 4301-4400 45 4401-4500 46 4501-4600 47 4601-4700 48 4701-4800 49 4801-4900 50 4901-5000 51 5001-5100 52 5101-5200 53 5201-5300 54 5301-5400 55 5401-5500 56 5501-5600 57 5601-5700 58 5701-5800 59 5801-5900 60 5901-6000 61 6001-6100 62 6101-6200 63 6201-6300 64 6301-6400 65 6401-6500 66 6501-6600 67 6601-6700 68 6701-6800 69 6801-6900 70 6901-7000 71 7001-7100 72 7101-7200 73 7201-7300 74 7301-7400 75 7401-7500 76 7501-7600 77 7601-7700 78 7701-7800 79 7801-7900 80 7901-8000 81 8001-8100 82 8101-8200 83 8201-8300 84 8301-8400 85 8401-8500 86 8501-8600 87 8601-8700 88 8701-8800 89 8801-8900 90 8901-9000 91 9001-9100 92 9101-9200 93 9201-9300 94 9301-9400 95 9401-9500 96 9501-9600 97 9601-9700 98 9701-9800 99 9801-9900 100 9901-10000 101 10001-10100 102 10101-10200 103 10201-10300 104 10301-10400 105 10401-10500 106 10501-10600 107 10601-10682

Color key:    Metamath Proof Explorer  Metamath Proof Explorer (1-8757)   Hilbert Space Explorer  Hilbert Space Explorer (8758-10682)  

Statement List for Metamath Proof Explorer - 4401-4500 - Page 45 of 107
TypeLabelDescription
Statement
 
Theoremdom2 4401 A mapping (first hypothesis) that is one-to-one (second hypothesis) implies its domain is dominated by its range. C and D can be read C(x) and D(y), as can be shown from their distinct variable conditions.
|- (x e. A -> C e. B)   &   |- ((x e. A /\ y e. A) -> (C = D <-> x = y))   =>   |- (A e. R -> A ~<_ B)
 
Theoremidssen 4402 Equality implies equinumerosity.
|- I (_ ~~
 
Theoremdmen 4403 The domain of equinumerosity.
|- dom ~~ = V
 
Theoremssdomg 4404 A set dominates its subsets. Theorem 16 of [Suppes] p. 94.
|- (A e. C -> (A (_ B -> A ~<_ B))
 
Theoremssdom2g 4405 A set dominates its subsets. Theorem 16 of [Suppes] p. 94.
|- (B e. C -> (A (_ B -> A ~<_ B))
 
Theoremener 4406 Equinumerosity is an equivalence relation.
|- Er ~~
 
Theoremensymg 4407 Symmetry of equinumerosity. Theorem 2 of [Suppes] p. 92.
|- (B e. C -> (A ~~ B -> B ~~ A))
 
Theoremensym 4408 Symmetry of equinumerosity. Theorem 2 of [Suppes] p. 92.
|- B e. V   =>   |- (A ~~ B -> B ~~ A)
 
Theoremensymi 4409 Symmetry of equinumerosity. Theorem 2 of [Suppes] p. 92.
|- B e. V   &   |- A ~~ B   =>   |- B ~~ A
 
Theorementrt 4410 Transitivity of equinumerosity. Theorem 3 of [Suppes] p. 92.
|- ((A ~~ B /\ B ~~ C) -> A ~~ C)
 
Theoremdomtr 4411 Transitivity of dominance relation. Theorem 17 of [Suppes] p. 94.
|- ((A ~<_ B /\ B ~<_ C) -> A ~<_ C)
 
Theorementr 4412 A chained equinumerosity inference.
|- A ~~ B   &   |- B ~~ C   =>   |- A ~~ C
 
Theorementr2 4413 A chained equinumerosity inference.
|- C e. V   &   |- A ~~ B   &   |- B ~~ C   =>   |- C ~~ A
 
Theorementr3 4414 A chained equinumerosity inference.
|- B e. V   &   |- A ~~ B   &   |- A ~~ C   =>   |- B ~~ C
 
Theorementr4 4415 A chained equinumerosity inference.
|- B e. V   &   |- A ~~ B   &   |- C ~~ B   =>   |- A ~~ C
 
Theoremendomtr 4416 Transitivity of equinumerosity and dominance.
|- ((A ~~ B /\ B ~<_ C) -> A ~<_ C)
 
Theoremdomentr 4417 Transitivity of dominance and equinumerosity.
|- ((A ~<_ B /\ B ~~ C) -> A ~<_ C)
 
Theoremf1imaen 4418 A one-to-one function's image under a subset of its domain is equinumerous to the subset.
|- C e. V   =>   |- ((F:A-1-1->B /\ C (_ A) -> (F"C) ~~ C)
 
Theoremen0 4419 The empty set is equinumerous only to itself. Exercise 1 of [TakeutiZaring] p. 88.
|- (A ~~ (/) <-> A = (/))
 
Theoremensn1 4420 A singleton is equinumerous to ordinal one.
|- A e. V   =>   |- {A} ~~ 1o
 
Theoremensn1g 4421 A singleton is equinumerous to ordinal one.
|- (A e. B -> {A} ~~ 1o)
 
Theoremen1 4422 A set is equinumerous to ordinal one iff it is a singleton.
|- (A ~~ 1o <-> E.x A = {x})
 
Theorem2dom 4423 A set that dominates ordinal 2 has at least 2 different members.
|- A e. V   =>   |- (2o ~<_ A -> E.x e. A E.y e. A -. x = y)
 
Theoremfundmen 4424 A function is equinumerous to its domain. Exercise 4 of [Suppes] p. 98.
|- F e. V   =>   |- (Fun F -> dom F ~~ F)
 
Theoremmapsnen 4425 Set exponentiation to a singleton exponent is equinumerous to its base. Exercise 4.43 of [Mendelson] p. 255.
|- A e. V   &   |- B e. V   =>   |- (A ^m {B}) ~~ A
 
Theoremmap1 4426 Set exponentiation: ordinal 1 to any set is equinumerous to ordinal 1. Exercise 4.42(b) of [Mendelson] p. 255.
|- A e. V   =>   |- (1o ^m A) ~~ 1o
 
Theoremen2sn 4427 Two singletons are equinumerous.
|- ((A e. C /\ B e. D) -> {A} ~~ {B})
 
Theoremsnfi 4428 A singleton is finite.
|- E.x e. om {A} ~~ x
 
Theoremunen 4429 Equinumerosity of union of disjoint sets. Theorem 4 of [Suppes] p. 92.
|- (((A ~~ B /\ C ~~ D) /\ ((A i^i C) = (/) /\ (B i^i D) = (/))) -> (A u. C) ~~ (B u. D))
 
Theoremxpsnen 4430 A set is equinumerous to its cross-product with a singleton. Proposition 4.22(c) of [Mendelson] p. 254.
|- A e. V   &   |- B e. V   =>   |- (A X. {B}) ~~ A
 
Theoremxpsneng 4431 A set is equinumerous to its cross-product with a singleton. Proposition 4.22(c) of [Mendelson] p. 254.
|- ((A e. C /\ B e. D) -> (A X. {B}) ~~ A)
 
Theoremendisj 4432 Any two sets are equinumerous to disjoint sets. Exercise 4.39 of [Mendelson] p. 255.
|- A e. V   &   |- B e. V   =>   |- E.xE.y((x ~~ A /\ y ~~ B) /\ (x i^i y) = (/))
 
Theoremundom 4433 Dominance law for union. Proposition 4.24(a) of [Mendelson] p. 257.
|- B e. V   &   |- C e. V   &   |- D e. V   =>   |- (((A ~<_ B /\ C ~<_ D) /\ (B i^i D) = (/)) -> (A u. C) ~<_ (B u. D))
 
Theoremxpcomen 4434 Commutative law for equinumerosity of cross product. Proposition 4.22(d) of [Mendelson] p. 254.
|- A e. V   &   |- B e. V   =>   |- (A X. B) ~~ (B X. A)
 
Theoremxpcomeng 4435 Commutative law for equinumerosity of cross product. Proposition 4.22(d) of [Mendelson] p. 254.
|- ((A e. C /\ B e. D) -> (A X. B) ~~ (B X. A))
 
Theoremxpassen 4436 Associative law for equinumerosity of cross product. Proposition 4.22(e) of [Mendelson] p. 254.
|- A e. V   &   |- B e. V   &   |- C e. V   =>   |- ((A X. B) X. C) ~~ (A X. (B X. C))
 
Theoremxpdom2 4437 Dominance law for cross product. Proposition 10.33(2) of [TakeutiZaring] p. 92.
|- B e. V   &   |- C e. V   =>   |- (A ~<_ B -> (C X. A) ~<_ (C X. B))
 
Theoremxpdom1 4438 Dominance law for cross product. Theorem 6L(c) of [Enderton] p. 149.
|- B e. V   &   |- C e. V   =>   |- (A ~<_ B -> (A X. C) ~<_ (B X. C))
 
Theoremxpdom1g 4439 Dominance law for cross product. Theorem 6L(c) of [Enderton] p. 149.
|- ((B e. R /\ C e. S /\ A ~<_ B) -> (A X. C) ~<_ (B X. C))
 
Theoremxpdom3 4440 A set is dominated by its cross product with a non-empty set. Exercise 6 of [Suppes] p. 98.
|- A e. V   =>   |- (B =/= (/) -> A ~<_ (A X. B))
 
Theorempw2en 4441 The power set of a set is equinumerous to set exponentiation with a base of ordinal 2. Proposition 10.44 of [TakeutiZaring] p. 96. (This proof seems excessively long. An attempt to find a shorter one is on my to-do list.)
|- A e. V   =>   |- P~A ~~ (2o ^m A)
 
Schroeder-Bernstein Theorem
 
Theoremsbthlem1 4442 Lemma for sbth 4452.
 
Theoremsbthlem2 4443 Lemma for sbth 4452.
 
Theoremsbthlem3 4444 Lemma for sbth 4452.
 
Theoremsbthlem4 4445 Lemma for sbth 4452.
 
Theoremsbthlem5 4446 Lemma for sbth 4452.
 
Theoremsbthlem6 4447 Lemma for sbth 4452.
 
Theoremsbthlem7 4448 Lemma for sbth 4452.
 
Theoremsbthlem8 4449 Lemma for sbth 4452.
 
Theoremsbthlem9 4450 Lemma for sbth 4452.
 
Theoremsbthlem10 4451 Lemma for sbth 4452.
 
Theoremsbth 4452 Schroeder-Bernstein Theorem. Theorem 18 of [Suppes] p. 95. This theorem states that if set A is smaller (has lower cardinality) than B and vice-versa, then A and B are equinumerous (have the same cardinality). The interesting thing is that this can be proved without invoking the Axiom of Choice, as we do here, but the proof as you can see is quite difficult. (The theorem can be proved more easily if we allow AC.) The main proof consists of lemmas sbthlem1 4442 through sbthlem10 4451; this final piece mainly changes bound variables to eliminate the hypotheses of sbthlem10 4451. We follow closely the proof in Suppes, which you should consult to understand our proof at a higher level.
|- ((A ~<_ B /\ B ~<_ A) -> A ~~ B)
 
Theoremsbthbg 4453 Schroeder-Bernstein Theorem and its converse.
|- (B e. C -> ((A ~<_ B /\ B ~<_ A) <-> A ~~ B))
 
Theoremsbthcl 4454 Schroeder-Bernstein Theorem in class form.
|- ~~ = ( ~<_ i^i `' ~<_ )
 
Theoremdfsdom2 4455 Alternate definition of strict dominance. Compare Definition 3 of [Suppes] p. 97.
|- ~< = ( ~<_ \ `' ~<_ )
 
Theorembrsdom2 4456 Alternate definition of strict dominance. Definition 3 of [Suppes] p. 97.
|- A e. V   &   |- B e. V   =>   |- (A ~< B <-> (A ~<_ B /\ -. B ~<_ A))
 
Theoremsdomnsym 4457 Strict dominance is not symmetric. Theorem 21(ii) of [Suppes] p. 97.
|- (A ~< B -> -. B ~< A)
 
Theoremdomnsym 4458 Theorem 22(i) of [Suppes] p. 97.
|- (A ~<_ B -> -. B ~< A)
 
Theorem0dom 4459 Any set dominates the empty set.
|- (/) ~<_ A
 
Theoremdom0 4460 A set dominated by the empty set is empty.
|- (A ~<_ (/) <-> A = (/))
 
Theorem0sdomg 4461 A set strictly dominates the empty set iff it is not empty.
|- (A e. B -> ((/) ~< A <-> A =/= (/)))
 
Theorem0sdom 4462 A set strictly dominates the empty set iff it is not empty.
|- A e. V   =>   |- ((/) ~< A <-> A =/= (/))
 
Theoremsdom0 4463 The empty set does not strictly dominate any set.
|- -. A ~< (/)
 
Theoremsdomdomtr 4464 Transitivity of strict dominance and dominance. Theorem 22(iii) of [Suppes] p. 97.
|- (C e. D -> ((A ~< B /\ B ~<_ C) -> A ~< C))
 
Theoremsdomentr 4465 Transitivity of strict dominance and equinumerosity. Exercise 11 of [Suppes] p. 98.
|- (C e. D -> ((A ~< B /\ B ~~ C) -> A ~< C))
 
Theoremensdomtr 4466 Transitivity of equinumerosity and strict dominance.
|- ((A ~~ B /\ B ~< C) -> A ~< C)
 
Theoremsdomirr 4467 Strict dominance is irreflexive. Theorem 21(i) of [Suppes] p. 97.
|- -. A ~< A
 
Theoremsdomex 4468 Technical lemma for simplifying proofs involving strict dominance.
|- (A ~< B -> (A e. V /\ B e. V))
 
Theoremsdomtr 4469 Strict dominance is transitive. Theorem 21(iii) of [Suppes] p. 97.
|- ((A ~< B /\ B ~< C) -> A ~< C)
 
Theoremsdomn2lp 4470 Strict dominance has no 2-cycle loops.
|- -. (A ~< B /\ B ~< A)
 
Theoremdomsdomtr 4471 Transitivity of dominance and strict dominance. Theorem 22(ii) of [Suppes] p. 97.
|- ((A ~<_ B /\ B ~< C) -> A ~< C)
 
Theoremenen1 4472 Equality-like theorem for equinumerosity.
|- ((B e. D /\ A ~~ B) -> (A ~~ C <-> B ~~ C))
 
Theoremenen2 4473 Equality-like theorem for equinumerosity.
|- ((B e. D /\ A ~~ B) -> (C ~~ A <-> C ~~ B))
 
Theoremdomen1 4474 Equality-like theorem for equinumerosity and dominance.
|- ((B e. D /\ A ~~ B) -> (A ~<_ C <-> B ~<_ C))
 
Theoremdomen2 4475 Equality-like theorem for equinumerosity and dominance.
|- ((B e. D /\ A ~~ B) -> (C ~<_ A <-> C ~<_ B))
 
Theoremsdomen1 4476 Equality-like theorem for equinumerosity and strict dominance.
|- ((B e. D /\ A ~~ B) -> (A ~< C <-> B ~< C))
 
Theoremsdomen2 4477 Equality-like theorem for equinumerosity and strict dominance.
|- ((B e. D /\ A ~~ B) -> (C ~< A <-> C ~< B))
 
Theoremfodomr 4478 There exists a mapping from a set onto any (non-empty) set that it dominates.
|- ((A e. C /\ (/) ~< B /\ B ~<_ A) -> E.f f:A-onto->B)
 
Theoremcanth2 4479 Cantor's Theorem. No set is equinumerous to its power set. Specifically, any set has a cardinality (size) strictly less than the cardinality of its power set. For example, the cardinality of real numbers is the same as the cardinality of the power set of integers, so real numbers cannot be put into a one-to-one correspondence with integers. Theorem 23 of [Suppes] p. 97. For the function version, see canth 3907.
|- A e. V   =>   |- A ~< P~A
 
Theorem2pwuninel 4480 The power set of the power set of the union of a set does not belong to the set. This theorem provides a way of constructing a new set that doesn't belong to a given set.
|- A e. V   =>   |- -. P~P~U.A e. A
 
Theoremcanth2g 4481 Cantor's theorem with the sethood requirement expressed as an antecedent. Theorem 23 of [Suppes] p. 97.
|- (A e. B -> A ~< P~A)
 
Theorempwuninel 4482 The power set of the union of a set does not belong to the set. This theorem provides a way of constructing a new set that doesn't belong to a given set.
|- -. P~U.A e. A
 
TheorempwuninelgOLD 4483 The power set of the union of a set does not belong to the set. This theorem provides a way of constructing a new set that doesn't belong to a given set.
|- (A e. B -> -. P~U.A e. A)
 
Theoremxpen 4484 Equinumerosity law for cross product. Proposition 4.22(b) of [Mendelson] p. 254.
|- A e. V   &   |- B e. V   &   |- C e. V