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Theorem List for Intuitionistic Logic Explorer - 3101-3200   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremssun2 3101 Subclass relationship for union of classes. (Contributed by NM, 30-Aug-1993.)
A ⊆ (BA)

Theoremssun3 3102 Subclass law for union of classes. (Contributed by NM, 5-Aug-1993.)
(ABA ⊆ (B𝐶))

Theoremssun4 3103 Subclass law for union of classes. (Contributed by NM, 14-Aug-1994.)
(ABA ⊆ (𝐶B))

Theoremelun1 3104 Membership law for union of classes. (Contributed by NM, 5-Aug-1993.)
(A BA (B𝐶))

Theoremelun2 3105 Membership law for union of classes. (Contributed by NM, 30-Aug-1993.)
(A BA (𝐶B))

Theoremunss1 3106 Subclass law for union of classes. (Contributed by NM, 14-Oct-1999.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
(AB → (A𝐶) ⊆ (B𝐶))

Theoremssequn1 3107 A relationship between subclass and union. Theorem 26 of [Suppes] p. 27. (Contributed by NM, 30-Aug-1993.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
(AB ↔ (AB) = B)

Theoremunss2 3108 Subclass law for union of classes. Exercise 7 of [TakeutiZaring] p. 18. (Contributed by NM, 14-Oct-1999.)
(AB → (𝐶A) ⊆ (𝐶B))

Theoremunss12 3109 Subclass law for union of classes. (Contributed by NM, 2-Jun-2004.)
((AB 𝐶𝐷) → (A𝐶) ⊆ (B𝐷))

Theoremssequn2 3110 A relationship between subclass and union. (Contributed by NM, 13-Jun-1994.)
(AB ↔ (BA) = B)

Theoremunss 3111 The union of two subclasses is a subclass. Theorem 27 of [Suppes] p. 27 and its converse. (Contributed by NM, 11-Jun-2004.)
((A𝐶 B𝐶) ↔ (AB) ⊆ 𝐶)

Theoremunssi 3112 An inference showing the union of two subclasses is a subclass. (Contributed by Raph Levien, 10-Dec-2002.)
A𝐶    &   B𝐶       (AB) ⊆ 𝐶

Theoremunssd 3113 A deduction showing the union of two subclasses is a subclass. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.)
(φA𝐶)    &   (φB𝐶)       (φ → (AB) ⊆ 𝐶)

Theoremunssad 3114 If (AB) is contained in 𝐶, so is A. One-way deduction form of unss 3111. Partial converse of unssd 3113. (Contributed by David Moews, 1-May-2017.)
(φ → (AB) ⊆ 𝐶)       (φA𝐶)

Theoremunssbd 3115 If (AB) is contained in 𝐶, so is B. One-way deduction form of unss 3111. Partial converse of unssd 3113. (Contributed by David Moews, 1-May-2017.)
(φ → (AB) ⊆ 𝐶)       (φB𝐶)

Theoremssun 3116 A condition that implies inclusion in the union of two classes. (Contributed by NM, 23-Nov-2003.)
((AB A𝐶) → A ⊆ (B𝐶))

Theoremrexun 3117 Restricted existential quantification over union. (Contributed by Jeff Madsen, 5-Jan-2011.)
(x (AB)φ ↔ (x A φ x B φ))

Theoremralunb 3118 Restricted quantification over a union. (Contributed by Scott Fenton, 12-Apr-2011.) (Proof shortened by Andrew Salmon, 29-Jun-2011.)
(x (AB)φ ↔ (x A φ x B φ))

Theoremralun 3119 Restricted quantification over union. (Contributed by Jeff Madsen, 2-Sep-2009.)
((x A φ x B φ) → x (AB)φ)

2.1.13.3  The intersection of two classes

Theoremelin 3120 Expansion of membership in an intersection of two classes. Theorem 12 of [Suppes] p. 25. (Contributed by NM, 29-Apr-1994.)
(A (B𝐶) ↔ (A B A 𝐶))

Theoremelin2 3121 Membership in a class defined as an intersection. (Contributed by Stefan O'Rear, 29-Mar-2015.)
𝑋 = (B𝐶)       (A 𝑋 ↔ (A B A 𝐶))

Theoremelin3 3122 Membership in a class defined as a ternary intersection. (Contributed by Stefan O'Rear, 29-Mar-2015.)
𝑋 = ((B𝐶) ∩ 𝐷)       (A 𝑋 ↔ (A B A 𝐶 A 𝐷))

Theoremincom 3123 Commutative law for intersection of classes. Exercise 7 of [TakeutiZaring] p. 17. (Contributed by NM, 5-Aug-1993.)
(AB) = (BA)

Theoremineqri 3124* Inference from membership to intersection. (Contributed by NM, 5-Aug-1993.)
((x A x B) ↔ x 𝐶)       (AB) = 𝐶

Theoremineq1 3125 Equality theorem for intersection of two classes. (Contributed by NM, 14-Dec-1993.)
(A = B → (A𝐶) = (B𝐶))

Theoremineq2 3126 Equality theorem for intersection of two classes. (Contributed by NM, 26-Dec-1993.)
(A = B → (𝐶A) = (𝐶B))

Theoremineq12 3127 Equality theorem for intersection of two classes. (Contributed by NM, 8-May-1994.)
((A = B 𝐶 = 𝐷) → (A𝐶) = (B𝐷))

Theoremineq1i 3128 Equality inference for intersection of two classes. (Contributed by NM, 26-Dec-1993.)
A = B       (A𝐶) = (B𝐶)

Theoremineq2i 3129 Equality inference for intersection of two classes. (Contributed by NM, 26-Dec-1993.)
A = B       (𝐶A) = (𝐶B)

Theoremineq12i 3130 Equality inference for intersection of two classes. (Contributed by NM, 24-Jun-2004.) (Proof shortened by Eric Schmidt, 26-Jan-2007.)
A = B    &   𝐶 = 𝐷       (A𝐶) = (B𝐷)

Theoremineq1d 3131 Equality deduction for intersection of two classes. (Contributed by NM, 10-Apr-1994.)
(φA = B)       (φ → (A𝐶) = (B𝐶))

Theoremineq2d 3132 Equality deduction for intersection of two classes. (Contributed by NM, 10-Apr-1994.)
(φA = B)       (φ → (𝐶A) = (𝐶B))

Theoremineq12d 3133 Equality deduction for intersection of two classes. (Contributed by NM, 24-Jun-2004.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
(φA = B)    &   (φ𝐶 = 𝐷)       (φ → (A𝐶) = (B𝐷))

Theoremineqan12d 3134 Equality deduction for intersection of two classes. (Contributed by NM, 7-Feb-2007.)
(φA = B)    &   (ψ𝐶 = 𝐷)       ((φ ψ) → (A𝐶) = (B𝐷))

Theoremdfss1 3135 A frequently-used variant of subclass definition df-ss 2925. (Contributed by NM, 10-Jan-2015.)
(AB ↔ (BA) = A)

Theoremdfss5 3136 Another definition of subclasshood. Similar to df-ss 2925, dfss 2926, and dfss1 3135. (Contributed by David Moews, 1-May-2017.)
(ABA = (BA))

Theoremnfin 3137 Bound-variable hypothesis builder for the intersection of classes. (Contributed by NM, 15-Sep-2003.) (Revised by Mario Carneiro, 14-Oct-2016.)
xA    &   xB       x(AB)

Theoremcsbing 3138 Distribute proper substitution through an intersection relation. (Contributed by Alan Sare, 22-Jul-2012.)
(A BA / x(𝐶𝐷) = (A / x𝐶A / x𝐷))

Theoremrabbi2dva 3139* Deduction from a wff to a restricted class abstraction. (Contributed by NM, 14-Jan-2014.)
((φ x A) → (x Bψ))       (φ → (AB) = {x Aψ})

Theoreminidm 3140 Idempotent law for intersection of classes. Theorem 15 of [Suppes] p. 26. (Contributed by NM, 5-Aug-1993.)
(AA) = A

Theoreminass 3141 Associative law for intersection of classes. Exercise 9 of [TakeutiZaring] p. 17. (Contributed by NM, 3-May-1994.)
((AB) ∩ 𝐶) = (A ∩ (B𝐶))

Theoremin12 3142 A rearrangement of intersection. (Contributed by NM, 21-Apr-2001.)
(A ∩ (B𝐶)) = (B ∩ (A𝐶))

Theoremin32 3143 A rearrangement of intersection. (Contributed by NM, 21-Apr-2001.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
((AB) ∩ 𝐶) = ((A𝐶) ∩ B)

Theoremin13 3144 A rearrangement of intersection. (Contributed by NM, 27-Aug-2012.)
(A ∩ (B𝐶)) = (𝐶 ∩ (BA))

Theoremin31 3145 A rearrangement of intersection. (Contributed by NM, 27-Aug-2012.)
((AB) ∩ 𝐶) = ((𝐶B) ∩ A)

Theoreminrot 3146 Rotate the intersection of 3 classes. (Contributed by NM, 27-Aug-2012.)
((AB) ∩ 𝐶) = ((𝐶A) ∩ B)

Theoremin4 3147 Rearrangement of intersection of 4 classes. (Contributed by NM, 21-Apr-2001.)
((AB) ∩ (𝐶𝐷)) = ((A𝐶) ∩ (B𝐷))

Theoreminindi 3148 Intersection distributes over itself. (Contributed by NM, 6-May-1994.)
(A ∩ (B𝐶)) = ((AB) ∩ (A𝐶))

Theoreminindir 3149 Intersection distributes over itself. (Contributed by NM, 17-Aug-2004.)
((AB) ∩ 𝐶) = ((A𝐶) ∩ (B𝐶))

Theoremsseqin2 3150 A relationship between subclass and intersection. Similar to Exercise 9 of [TakeutiZaring] p. 18. (Contributed by NM, 17-May-1994.)
(AB ↔ (BA) = A)

Theoreminss1 3151 The intersection of two classes is a subset of one of them. Part of Exercise 12 of [TakeutiZaring] p. 18. (Contributed by NM, 27-Apr-1994.)
(AB) ⊆ A

Theoreminss2 3152 The intersection of two classes is a subset of one of them. Part of Exercise 12 of [TakeutiZaring] p. 18. (Contributed by NM, 27-Apr-1994.)
(AB) ⊆ B

Theoremssin 3153 Subclass of intersection. Theorem 2.8(vii) of [Monk1] p. 26. (Contributed by NM, 15-Jun-2004.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
((AB A𝐶) ↔ A ⊆ (B𝐶))

Theoremssini 3154 An inference showing that a subclass of two classes is a subclass of their intersection. (Contributed by NM, 24-Nov-2003.)
AB    &   A𝐶       A ⊆ (B𝐶)

Theoremssind 3155 A deduction showing that a subclass of two classes is a subclass of their intersection. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.)
(φAB)    &   (φA𝐶)       (φA ⊆ (B𝐶))

Theoremssrin 3156 Add right intersection to subclass relation. (Contributed by NM, 16-Aug-1994.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
(AB → (A𝐶) ⊆ (B𝐶))

Theoremsslin 3157 Add left intersection to subclass relation. (Contributed by NM, 19-Oct-1999.)
(AB → (𝐶A) ⊆ (𝐶B))

Theoremss2in 3158 Intersection of subclasses. (Contributed by NM, 5-May-2000.)
((AB 𝐶𝐷) → (A𝐶) ⊆ (B𝐷))

Theoremssinss1 3159 Intersection preserves subclass relationship. (Contributed by NM, 14-Sep-1999.)
(A𝐶 → (AB) ⊆ 𝐶)

Theoreminss 3160 Inclusion of an intersection of two classes. (Contributed by NM, 30-Oct-2014.)
((A𝐶 B𝐶) → (AB) ⊆ 𝐶)

2.1.13.4  Combinations of difference, union, and intersection of two classes

Theoremunabs 3161 Absorption law for union. (Contributed by NM, 16-Apr-2006.)
(A ∪ (AB)) = A

Theoreminabs 3162 Absorption law for intersection. (Contributed by NM, 16-Apr-2006.)
(A ∩ (AB)) = A

Theoremnssinpss 3163 Negation of subclass expressed in terms of intersection and proper subclass. (Contributed by NM, 30-Jun-2004.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
AB ↔ (AB) ⊊ A)

Theoremnsspssun 3164 Negation of subclass expressed in terms of proper subclass and union. (Contributed by NM, 15-Sep-2004.)
ABB ⊊ (AB))

Theoremssddif 3165 Double complement and subset. Similar to ddifss 3169 but inside a class B instead of the universal class V. In classical logic the subset operation on the right hand side could be an equality (that is, AB ↔ (B ∖ (BA)) = A). (Contributed by Jim Kingdon, 24-Jul-2018.)
(ABA ⊆ (B ∖ (BA)))

Theoremunssdif 3166 Union of two classes and class difference. In classical logic this would be an equality. (Contributed by Jim Kingdon, 24-Jul-2018.)
(AB) ⊆ (V ∖ ((V ∖ A) ∖ B))

Theoreminssdif 3167 Intersection of two classes and class difference. In classical logic this would be an equality. (Contributed by Jim Kingdon, 24-Jul-2018.)
(AB) ⊆ (A ∖ (V ∖ B))

Theoremdifin 3168 Difference with intersection. Theorem 33 of [Suppes] p. 29. (Contributed by NM, 31-Mar-1998.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
(A ∖ (AB)) = (AB)

Theoremddifss 3169 Double complement under universal class. In classical logic (or given an additional hypothesis, as in ddifnel 3069), this is equality rather than subset. (Contributed by Jim Kingdon, 24-Jul-2018.)
A ⊆ (V ∖ (V ∖ A))

Theoremunssin 3170 Union as a subset of class complement and intersection (De Morgan's law). One direction of the definition of union in [Mendelson] p. 231. This would be an equality, rather than subset, in classical logic. (Contributed by Jim Kingdon, 25-Jul-2018.)
(AB) ⊆ (V ∖ ((V ∖ A) ∩ (V ∖ B)))

Theoreminssun 3171 Intersection in terms of class difference and union (De Morgan's law). Similar to Exercise 4.10(n) of [Mendelson] p. 231. This would be an equality, rather than subset, in classical logic. (Contributed by Jim Kingdon, 25-Jul-2018.)
(AB) ⊆ (V ∖ ((V ∖ A) ∪ (V ∖ B)))

Theoreminssddif 3172 Intersection of two classes and class difference. In classical logic, such as Exercise 4.10(q) of [Mendelson] p. 231, this is an equality rather than subset. (Contributed by Jim Kingdon, 26-Jul-2018.)
(AB) ⊆ (A ∖ (AB))

Theoreminvdif 3173 Intersection with universal complement. Remark in [Stoll] p. 20. (Contributed by NM, 17-Aug-2004.)
(A ∩ (V ∖ B)) = (AB)

Theoremindif 3174 Intersection with class difference. Theorem 34 of [Suppes] p. 29. (Contributed by NM, 17-Aug-2004.)
(A ∩ (AB)) = (AB)

Theoremindif2 3175 Bring an intersection in and out of a class difference. (Contributed by Jeff Hankins, 15-Jul-2009.)
(A ∩ (B𝐶)) = ((AB) ∖ 𝐶)

Theoremindif1 3176 Bring an intersection in and out of a class difference. (Contributed by Mario Carneiro, 15-May-2015.)
((A𝐶) ∩ B) = ((AB) ∖ 𝐶)

Theoremindifcom 3177 Commutation law for intersection and difference. (Contributed by Scott Fenton, 18-Feb-2013.)
(A ∩ (B𝐶)) = (B ∩ (A𝐶))

Theoremindi 3178 Distributive law for intersection over union. Exercise 10 of [TakeutiZaring] p. 17. (Contributed by NM, 30-Sep-2002.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
(A ∩ (B𝐶)) = ((AB) ∪ (A𝐶))

Theoremundi 3179 Distributive law for union over intersection. Exercise 11 of [TakeutiZaring] p. 17. (Contributed by NM, 30-Sep-2002.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
(A ∪ (B𝐶)) = ((AB) ∩ (A𝐶))

Theoremindir 3180 Distributive law for intersection over union. Theorem 28 of [Suppes] p. 27. (Contributed by NM, 30-Sep-2002.)
((AB) ∩ 𝐶) = ((A𝐶) ∪ (B𝐶))

Theoremundir 3181 Distributive law for union over intersection. Theorem 29 of [Suppes] p. 27. (Contributed by NM, 30-Sep-2002.)
((AB) ∪ 𝐶) = ((A𝐶) ∩ (B𝐶))

Theoremuneqin 3182 Equality of union and intersection implies equality of their arguments. (Contributed by NM, 16-Apr-2006.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
((AB) = (AB) ↔ A = B)

Theoremdifundi 3183 Distributive law for class difference. Theorem 39 of [Suppes] p. 29. (Contributed by NM, 17-Aug-2004.)
(A ∖ (B𝐶)) = ((AB) ∩ (A𝐶))

Theoremdifundir 3184 Distributive law for class difference. (Contributed by NM, 17-Aug-2004.)
((AB) ∖ 𝐶) = ((A𝐶) ∪ (B𝐶))

Theoremdifindiss 3185 Distributive law for class difference. In classical logic, for example, theorem 40 of [Suppes] p. 29, this is an equality instead of subset. (Contributed by Jim Kingdon, 26-Jul-2018.)
((AB) ∪ (A𝐶)) ⊆ (A ∖ (B𝐶))

Theoremdifindir 3186 Distributive law for class difference. (Contributed by NM, 17-Aug-2004.)
((AB) ∖ 𝐶) = ((A𝐶) ∩ (B𝐶))

Theoremindifdir 3187 Distribute intersection over difference. (Contributed by Scott Fenton, 14-Apr-2011.)
((AB) ∩ 𝐶) = ((A𝐶) ∖ (B𝐶))

Theoremdifdif2ss 3188 Set difference with a set difference. In classical logic this would be equality rather than subset. (Contributed by Jim Kingdon, 27-Jul-2018.)
((AB) ∪ (A𝐶)) ⊆ (A ∖ (B𝐶))

Theoremundm 3189 De Morgan's law for union. Theorem 5.2(13) of [Stoll] p. 19. (Contributed by NM, 18-Aug-2004.)
(V ∖ (AB)) = ((V ∖ A) ∩ (V ∖ B))

Theoremindmss 3190 De Morgan's law for intersection. In classical logic, this would be equality rather than subset, as in Theorem 5.2(13') of [Stoll] p. 19. (Contributed by Jim Kingdon, 27-Jul-2018.)
((V ∖ A) ∪ (V ∖ B)) ⊆ (V ∖ (AB))

Theoremdifun1 3191 A relationship involving double difference and union. (Contributed by NM, 29-Aug-2004.)
(A ∖ (B𝐶)) = ((AB) ∖ 𝐶)

Theoremundif3ss 3192 A subset relationship involving class union and class difference. In classical logic, this would be equality rather than subset, as in the first equality of Exercise 13 of [TakeutiZaring] p. 22. (Contributed by Jim Kingdon, 28-Jul-2018.)
(A ∪ (B𝐶)) ⊆ ((AB) ∖ (𝐶A))

Theoremdifin2 3193 Represent a set difference as an intersection with a larger difference. (Contributed by Jeff Madsen, 2-Sep-2009.)
(A𝐶 → (AB) = ((𝐶B) ∩ A))

Theoremdif32 3194 Swap second and third argument of double difference. (Contributed by NM, 18-Aug-2004.)
((AB) ∖ 𝐶) = ((A𝐶) ∖ B)

Theoremdifabs 3195 Absorption-like law for class difference: you can remove a class only once. (Contributed by FL, 2-Aug-2009.)
((AB) ∖ B) = (AB)

Theoremsymdif1 3196 Two ways to express symmetric difference. This theorem shows the equivalence of the definition of symmetric difference in [Stoll] p. 13 and the restated definition in Example 4.1 of [Stoll] p. 262. (Contributed by NM, 17-Aug-2004.)
((AB) ∪ (BA)) = ((AB) ∖ (AB))

2.1.13.5  Class abstractions with difference, union, and intersection of two classes

Theoremsymdifxor 3197* Expressing symmetric difference with exclusive-or or two differences. (Contributed by Jim Kingdon, 28-Jul-2018.)
((AB) ∪ (BA)) = {x ∣ (x Ax B)}

Theoremunab 3198 Union of two class abstractions. (Contributed by NM, 29-Sep-2002.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
({xφ} ∪ {xψ}) = {x ∣ (φ ψ)}

Theoreminab 3199 Intersection of two class abstractions. (Contributed by NM, 29-Sep-2002.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
({xφ} ∩ {xψ}) = {x ∣ (φ ψ)}

Theoremdifab 3200 Difference of two class abstractions. (Contributed by NM, 23-Oct-2004.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
({xφ} ∖ {xψ}) = {x ∣ (φ ¬ ψ)}

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