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Statement List for Metamath Proof Explorer - 3301-3400 - Page 34 of 58
TypeLabelDescription
Statement
 
Theoremf1oen 3301 The domain and range of a one-to-one, onto function are equinumerous.
|- A e. V   =>   |- (F:A-1-1-onto->B -> A ~~ B)
 
Theoremf1dom 3302 The domain of a one-to-one function is dominated by its range.
|- A e. V   =>   |- (F:A-1-1->B -> A ~<_ B)
 
Theoremen2d 3303 Equinumerosity inference from an implicit one-to-one onto function.
|- (ph -> A e. V)   &   |- (ph -> (x e. A -> C e. V))   &   |- (ph -> (y e. B -> D e. V))   &   |- (ph -> ((x e. A /\ y = C) <-> (y e. B /\ x = D)))   =>   |- (ph -> A ~~ B)
 
Theoremen3d 3304 Equinumerosity inference from an implicit one-to-one onto function.
|- (ph -> A e. V)   &   |- (ph -> (x e. A -> C e. B))   &   |- (ph -> (y e. B -> D e. A))   &   |- (ph -> ((x e. A /\ y e. B) -> (x = D <-> y = C)))   =>   |- (ph -> A ~~ B)
 
Theoremen2 3305 Equinumerosity inference from an implicit one-to-one onto function.
|- A e. V   &   |- (x e. A -> C e. V)   &   |- (y e. B -> D e. V)   &   |- ((x e. A /\ y = C) <-> (y e. B /\ x = D))   =>   |- A ~~ B
 
Theoremen3 3306 Equinumerosity inference from an implicit one-to-one onto function.
|- A e. V   &   |- (x e. A -> C e. B)   &   |- (y e. B -> D e. A)   &   |- ((x e. A /\ y e. B) -> (x = D <-> y = C))   =>   |- A ~~ B
 
Theoremdom2d 3307 A mapping (first hypothesis) that is one-to-one (second hypothesis) implies its domain is dominated by its range.
|- (ph -> (x e. A -> C e. B))   &   |- (ph -> ((x e. A /\ y e. A) -> (C = D <-> x = y)))   =>   |- (ph -> (A e. R -> A ~<_ B))
 
Theoremdom2 3308 A mapping (first hypothesis) that is one-to-one (second hypothesis) implies its domain is dominated by its range.
|- (x e. A -> C e. B)   &   |- ((x e. A /\ y e. A) -> (C = D <-> x = y))   =>   |- (A e. R -> A ~<_ B)
 
Theoremidssen 3309 Equality implies equinumerosity.
|- I (_ ~~
 
Theoremdmen 3310 The domain of equinumerosity.
|- dom ~~ = V
 
Theoremssdomg 3311 A set dominates its subsets. Theorem 16 of [Suppes] p. 94.
|- (A e. C -> (A (_ B -> A ~<_ B))
 
Theoremssdom2g 3312 A set dominates its subsets. Theorem 16 of [Suppes] p. 94.
|- (B e. C -> (A (_ B -> A ~<_ B))
 
Theoremener 3313 Equinumerosity is an equivalence relation.
|- Er ~~
 
Theoremsdomirr 3314 Strict dominance is irreflexive. Theorem 21(i) of [Suppes] p. 97.
|- -. A ~< A
 
Theoremsdomex 3315 Technical lemma for simplifying proofs involving strict dominance.
|- (A ~< B -> (A e. V /\ B e. V))
 
Theoremensymg 3316 Symmetry of equinumerosity. Theorem 2 of [Suppes] p. 92.
|- (B e. C -> (A ~~ B -> B ~~ A))
 
Theoremensym 3317 Symmetry of equinumerosity. Theorem 2 of [Suppes] p. 92.
|- B e. V   =>   |- (A ~~ B -> B ~~ A)
 
Theoremensymi 3318 Symmetry of equinumerosity. Theorem 2 of [Suppes] p. 92.
|- B e. V   &   |- A ~~ B   =>   |- B ~~ A
 
Theorementrt 3319 Transitivity of equinumerosity. Theorem 3 of [Suppes] p. 92.
|- ((A ~~ B /\ B ~~ C) -> A ~~ C)
 
Theoremdomtr 3320 Transitivity of dominance relation. Theorem 17 of [Suppes] p. 94.
|- ((A ~<_ B /\ B ~<_ C) -> A ~<_ C)
 
Theorementr 3321 A chained equinumerosity inference.
|- A ~~ B   &   |- B ~~ C   =>   |- A ~~ C
 
Theorementr2 3322 A chained equinumerosity inference.
|- C e. V   &   |- A ~~ B   &   |- B ~~ C   =>   |- C ~~ A
 
Theorementr3 3323 A chained equinumerosity inference.
|- B e. V   &   |- A ~~ B   &   |- A ~~ C   =>   |- B ~~ C
 
Theorementr4 3324 A chained equinumerosity inference.
|- B e. V   &   |- A ~~ B   &   |- C ~~ B   =>   |- A ~~ C
 
Theoremendomtr 3325 Transitivity of equinumerosity and dominance.
|- ((A ~~ B /\ B ~<_ C) -> A ~<_ C)
 
Theoremdomentr 3326 Transitivity of dominance and equinumerosity.
|- ((A ~<_ B /\ B ~~ C) -> A ~<_ C)
 
Theoremf1imaen 3327 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 3328 The empty set is equinumerous only to itself. Exercise 1 of [TakeutiZaring] p. 88.
|- (A ~~ (/) <-> A = (/))
 
Theoremensn1 3329 A singleton is equinumerous to ordinal one.
|- A e. V   =>   |- {A} ~~ 1o
 
Theoremensn1g 3330 A singleton is equinumerous to ordinal one.
|- (A e. B -> {A} ~~ 1o)
 
Theoremen1 3331 A set is equinumerous to ordinal one iff it is a singleton.
|- (A ~~ 1o <-> E.x A = {x})
 
Theorem2dom 3332 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 3333 A function is equinumerous to its domain. Exercise 4 of [Suppes] p. 98.
|- F e. V   =>   |- (Fun F -> dom F ~~ F)
 
Theoremmapsnen 3334 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 3335 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 3336 Two singletons are equinumerous.
|- ((A e. C /\ B e. D) -> {A} ~~ {B})
 
Theoremsnfi 3337 A singleton is finite.
|- E.x e. om {A} ~~ x
 
Theoremunen 3338 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 3339 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 3340 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 3341 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 3342 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 3343 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)
 
Theoremxpassen 3344 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 3345 Dominance law for cross product. Proposition 10.33(2) of [TakeutiZaring] p. 92.
|- A e. V   &   |- B e. V   &   |- C e. V   =>   |- (A ~<_ B -> (C X. A) ~<_ (C X. B))
 
Theoremxpdom1 3346 Dominance law for cross product. Theorem 6L(c) of [Enderton] p. 149.
|- A e. V   &   |- B e. V   &   |- C e. V   =>   |- (A ~<_ B -> (A X. C) ~<_ (B X. C))
 
Theoremxpdom3 3347 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 3348 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)
 
Theoremsbthlem1 3349 Lemma for Schroeder-Bernstein Theorem.
|- A e. V   &   |- D = {x | (x (_ A /\ (g"(B \ (f"x))) (_ (A \ x))}   =>   |- U.D (_ (A \ (g"(B \ (f"U.D))))
 
Theoremsbthlem2 3350 Lemma for Schroeder-Bernstein Theorem.
|- A e. V   &   |- D = {x | (x (_ A /\ (g"(B \ (f"x))) (_ (A \ x))}   =>   |- (ran g (_ A -> (A \ (g"(B \ (f"U.D)))) (_ U.D)
 
Theoremsbthlem3 3351 Lemma for Schroeder-Bernstein Theorem.
|- A e. V   &   |- D = {x | (x (_ A /\ (g"(B \ (f"x))) (_ (A \ x))}   =>   |- (ran g (_ A -> (g"(B \ (f"U.D))) = (A \ U.D))
 
Theoremsbthlem4 3352 Lemma for Schroeder-Bernstein Theorem.
|- A e. V   &   |- D = {x | (x (_ A /\ (g"(B \ (f"x))) (_ (A \ x))}   =>   |- (((dom g = B /\ ran g (_ A) /\ Fun `'g) -> (`'g"(A \ U.D)) = (B \ (f"U.D)))
 
Theoremsbthlem5 3353 Lemma for Schroeder-Bernstein Theorem.
|- A e. V   &   |- D = {x | (x (_ A /\ (g"(B \ (f"x))) (_ (A \ x))}   &   |- H = ((f |` U.D) u. (`'g |` (A \ U.D)))   =>   |- ((dom f = A /\ ran g (_ A) -> dom H = A)
 
Theoremsbthlem6 3354 Lemma for Schroeder-Bernstein Theorem.
|- A e. V   &   |- D = {x | (x (_ A /\ (g"(B \ (f"x))) (_ (A \ x))}   &   |- H = ((f |` U.D) u. (`'g |` (A \ U.D)))   =>   |- ((ran f (_ B /\ ((dom g = B /\ ran g (_ A) /\ Fun `'g)) -> ran H = B)
 
Theoremsbthlem7 3355 Lemma for Schroeder-Bernstein Theorem.
|- A e. V   &   |- D = {x | (x (_ A /\ (g"(B \ (f"x))) (_ (A \ x))}   &   |- H = ((f |` U.D) u. (`'g |` (A \ U.D)))   =>   |- ((Fun f /\ Fun `'g) <