 Home Intuitionistic Logic ExplorerTheorem List (p. 30 of 74) < Previous  Next > Bad symbols? Try the GIF version. Mirrors  >  Metamath Home Page  >  ILE Home Page  >  Theorem List Contents  >  Recent Proofs       This page: Page List

Theorem List for Intuitionistic Logic Explorer - 2901-3000   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremeldifd 2901 If a class is in one class and not another, it is also in their difference. One-way deduction form of eldif 2900. (Contributed by David Moews, 1-May-2017.)
(φA B)    &   (φ → ¬ A 𝐶)       (φA (B𝐶))

Theoremeldifad 2902 If a class is in the difference of two classes, it is also in the minuend. One-way deduction form of eldif 2900. (Contributed by David Moews, 1-May-2017.)
(φA (B𝐶))       (φA B)

Theoremeldifbd 2903 If a class is in the difference of two classes, it is not in the subtrahend. One-way deduction form of eldif 2900. (Contributed by David Moews, 1-May-2017.)
(φA (B𝐶))       (φ → ¬ A 𝐶)

2.1.12  Subclasses and subsets

Definitiondf-ss 2904 Define the subclass relationship. Exercise 9 of [TakeutiZaring] p. 18. Note that AA (proved in ssid 2937). Contrast this relationship with the relationship AB (as will be defined in df-pss 2906). For a more traditional definition, but requiring a dummy variable, see dfss2 2907 (or dfss3 2908 which is similar). (Contributed by NM, 27-Apr-1994.)
(AB ↔ (AB) = A)

Theoremdfss 2905 Variant of subclass definition df-ss 2904. (Contributed by NM, 3-Sep-2004.)
(ABA = (AB))

Definitiondf-pss 2906 Define proper subclass relationship between two classes. Definition 5.9 of [TakeutiZaring] p. 17. Note that ¬ AA (proved in pssirr 3017). Contrast this relationship with the relationship AB (as defined in df-ss 2904). Other possible definitions are given by dfpss2 3002 and dfpss3 3003. (Contributed by NM, 7-Feb-1996.)
(AB ↔ (AB AB))

Theoremdfss2 2907* Alternate definition of the subclass relationship between two classes. Definition 5.9 of [TakeutiZaring] p. 17. (Contributed by NM, 8-Jan-2002.)
(ABx(x Ax B))

Theoremdfss3 2908* Alternate definition of subclass relationship. (Contributed by NM, 14-Oct-1999.)
(ABx A x B)

Theoremdfss2f 2909 Equivalence for subclass relation, using bound-variable hypotheses instead of distinct variable conditions. (Contributed by NM, 3-Jul-1994.) (Revised by Andrew Salmon, 27-Aug-2011.)
xA    &   xB       (ABx(x Ax B))

Theoremdfss3f 2910 Equivalence for subclass relation, using bound-variable hypotheses instead of distinct variable conditions. (Contributed by NM, 20-Mar-2004.)
xA    &   xB       (ABx A x B)

Theoremnfss 2911 If x is not free in A and B, it is not free in AB. (Contributed by NM, 27-Dec-1996.)
xA    &   xB       x AB

Theoremssel 2912 Membership relationships follow from a subclass relationship. (Contributed by NM, 5-Aug-1993.)
(AB → (𝐶 A𝐶 B))

Theoremssel2 2913 Membership relationships follow from a subclass relationship. (Contributed by NM, 7-Jun-2004.)
((AB 𝐶 A) → 𝐶 B)

Theoremsseli 2914 Membership inference from subclass relationship. (Contributed by NM, 5-Aug-1993.)
AB       (𝐶 A𝐶 B)

Theoremsselii 2915 Membership inference from subclass relationship. (Contributed by NM, 31-May-1999.)
AB    &   𝐶 A       𝐶 B

Theoremsseldi 2916 Membership inference from subclass relationship. (Contributed by NM, 25-Jun-2014.)
AB    &   (φ𝐶 A)       (φ𝐶 B)

Theoremsseld 2917 Membership deduction from subclass relationship. (Contributed by NM, 15-Nov-1995.)
(φAB)       (φ → (𝐶 A𝐶 B))

Theoremsselda 2918 Membership deduction from subclass relationship. (Contributed by NM, 26-Jun-2014.)
(φAB)       ((φ 𝐶 A) → 𝐶 B)

Theoremsseldd 2919 Membership inference from subclass relationship. (Contributed by NM, 14-Dec-2004.)
(φAB)    &   (φ𝐶 A)       (φ𝐶 B)

Theoremssneld 2920 If a class is not in another class, it is also not in a subclass of that class. Deduction form. (Contributed by David Moews, 1-May-2017.)
(φAB)       (φ → (¬ 𝐶 B → ¬ 𝐶 A))

Theoremssneldd 2921 If an element is not in a class, it is also not in a subclass of that class. Deduction form. (Contributed by David Moews, 1-May-2017.)
(φAB)    &   (φ → ¬ 𝐶 B)       (φ → ¬ 𝐶 A)

Theoremssriv 2922* Inference rule based on subclass definition. (Contributed by NM, 5-Aug-1993.)
(x Ax B)       AB

Theoremssrd 2923 Deduction rule based on subclass definition. (Contributed by Thierry Arnoux, 8-Mar-2017.)
xφ    &   xA    &   xB    &   (φ → (x Ax B))       (φAB)

Theoremssrdv 2924* Deduction rule based on subclass definition. (Contributed by NM, 15-Nov-1995.)
(φ → (x Ax B))       (φAB)

Theoremsstr2 2925 Transitivity of subclasses. Exercise 5 of [TakeutiZaring] p. 17. (Contributed by NM, 5-Aug-1993.) (Proof shortened by Andrew Salmon, 14-Jun-2011.)
(AB → (B𝐶A𝐶))

Theoremsstr 2926 Transitivity of subclasses. Theorem 6 of [Suppes] p. 23. (Contributed by NM, 5-Sep-2003.)
((AB B𝐶) → A𝐶)

Theoremsstri 2927 Subclass transitivity inference. (Contributed by NM, 5-May-2000.)
AB    &   B𝐶       A𝐶

Theoremsstrd 2928 Subclass transitivity deduction. (Contributed by NM, 2-Jun-2004.)
(φAB)    &   (φB𝐶)       (φA𝐶)

Theoremsyl5ss 2929 Subclass transitivity deduction. (Contributed by NM, 6-Feb-2014.)
AB    &   (φB𝐶)       (φA𝐶)

Theoremsyl6ss 2930 Subclass transitivity deduction. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.)
(φAB)    &   B𝐶       (φA𝐶)

Theoremsylan9ss 2931 A subclass transitivity deduction. (Contributed by NM, 27-Sep-2004.) (Proof shortened by Andrew Salmon, 14-Jun-2011.)
(φAB)    &   (ψB𝐶)       ((φ ψ) → A𝐶)

Theoremsylan9ssr 2932 A subclass transitivity deduction. (Contributed by NM, 27-Sep-2004.)
(φAB)    &   (ψB𝐶)       ((ψ φ) → A𝐶)

Theoremeqss 2933 The subclass relationship is antisymmetric. Compare Theorem 4 of [Suppes] p. 22. (Contributed by NM, 5-Aug-1993.)
(A = B ↔ (AB BA))

Theoremeqssi 2934 Infer equality from two subclass relationships. Compare Theorem 4 of [Suppes] p. 22. (Contributed by NM, 9-Sep-1993.)
AB    &   BA       A = B

Theoremeqssd 2935 Equality deduction from two subclass relationships. Compare Theorem 4 of [Suppes] p. 22. (Contributed by NM, 27-Jun-2004.)
(φAB)    &   (φBA)       (φA = B)

Theoremeqrd 2936 Deduce equality of classes from equivalence of membership. (Contributed by Thierry Arnoux, 21-Mar-2017.)
xφ    &   xA    &   xB    &   (φ → (x Ax B))       (φA = B)

Theoremssid 2937 Any class is a subclass of itself. Exercise 10 of [TakeutiZaring] p. 18. (Contributed by NM, 5-Aug-1993.) (Proof shortened by Andrew Salmon, 14-Jun-2011.)
AA

Theoremssv 2938 Any class is a subclass of the universal class. (Contributed by NM, 31-Oct-1995.)
A ⊆ V

Theoremsseq1 2939 Equality theorem for subclasses. (Contributed by NM, 5-Aug-1993.) (Proof shortened by Andrew Salmon, 21-Jun-2011.)
(A = B → (A𝐶B𝐶))

Theoremsseq2 2940 Equality theorem for the subclass relationship. (Contributed by NM, 25-Jun-1998.)
(A = B → (𝐶A𝐶B))

Theoremsseq12 2941 Equality theorem for the subclass relationship. (Contributed by NM, 31-May-1999.)
((A = B 𝐶 = 𝐷) → (A𝐶B𝐷))

Theoremsseq1i 2942 An equality inference for the subclass relationship. (Contributed by NM, 18-Aug-1993.)
A = B       (A𝐶B𝐶)

Theoremsseq2i 2943 An equality inference for the subclass relationship. (Contributed by NM, 30-Aug-1993.)
A = B       (𝐶A𝐶B)

Theoremsseq12i 2944 An equality inference for the subclass relationship. (Contributed by NM, 31-May-1999.) (Proof shortened by Eric Schmidt, 26-Jan-2007.)
A = B    &   𝐶 = 𝐷       (A𝐶B𝐷)

Theoremsseq1d 2945 An equality deduction for the subclass relationship. (Contributed by NM, 14-Aug-1994.)
(φA = B)       (φ → (A𝐶B𝐶))

Theoremsseq2d 2946 An equality deduction for the subclass relationship. (Contributed by NM, 14-Aug-1994.)
(φA = B)       (φ → (𝐶A𝐶B))

Theoremsseq12d 2947 An equality deduction for the subclass relationship. (Contributed by NM, 31-May-1999.)
(φA = B)    &   (φ𝐶 = 𝐷)       (φ → (A𝐶B𝐷))

Theoremeqsstri 2948 Substitution of equality into a subclass relationship. (Contributed by NM, 16-Jul-1995.)
A = B    &   B𝐶       A𝐶

Theoremeqsstr3i 2949 Substitution of equality into a subclass relationship. (Contributed by NM, 19-Oct-1999.)
B = A    &   B𝐶       A𝐶

Theoremsseqtri 2950 Substitution of equality into a subclass relationship. (Contributed by NM, 28-Jul-1995.)
AB    &   B = 𝐶       A𝐶

Theoremsseqtr4i 2951 Substitution of equality into a subclass relationship. (Contributed by NM, 4-Apr-1995.)
AB    &   𝐶 = B       A𝐶

Theoremeqsstrd 2952 Substitution of equality into a subclass relationship. (Contributed by NM, 25-Apr-2004.)
(φA = B)    &   (φB𝐶)       (φA𝐶)

Theoremeqsstr3d 2953 Substitution of equality into a subclass relationship. (Contributed by NM, 25-Apr-2004.)
(φB = A)    &   (φB𝐶)       (φA𝐶)

Theoremsseqtrd 2954 Substitution of equality into a subclass relationship. (Contributed by NM, 25-Apr-2004.)
(φAB)    &   (φB = 𝐶)       (φA𝐶)

Theoremsseqtr4d 2955 Substitution of equality into a subclass relationship. (Contributed by NM, 25-Apr-2004.)
(φAB)    &   (φ𝐶 = B)       (φA𝐶)

Theorem3sstr3i 2956 Substitution of equality in both sides of a subclass relationship. (Contributed by NM, 13-Jan-1996.) (Proof shortened by Eric Schmidt, 26-Jan-2007.)
AB    &   A = 𝐶    &   B = 𝐷       𝐶𝐷

Theorem3sstr4i 2957 Substitution of equality in both sides of a subclass relationship. (Contributed by NM, 13-Jan-1996.) (Proof shortened by Eric Schmidt, 26-Jan-2007.)
AB    &   𝐶 = A    &   𝐷 = B       𝐶𝐷

Theorem3sstr3g 2958 Substitution of equality into both sides of a subclass relationship. (Contributed by NM, 1-Oct-2000.)
(φAB)    &   A = 𝐶    &   B = 𝐷       (φ𝐶𝐷)

Theorem3sstr4g 2959 Substitution of equality into both sides of a subclass relationship. (Contributed by NM, 16-Aug-1994.) (Proof shortened by Eric Schmidt, 26-Jan-2007.)
(φAB)    &   𝐶 = A    &   𝐷 = B       (φ𝐶𝐷)

Theorem3sstr3d 2960 Substitution of equality into both sides of a subclass relationship. (Contributed by NM, 1-Oct-2000.)
(φAB)    &   (φA = 𝐶)    &   (φB = 𝐷)       (φ𝐶𝐷)

Theorem3sstr4d 2961 Substitution of equality into both sides of a subclass relationship. (Contributed by NM, 30-Nov-1995.) (Proof shortened by Eric Schmidt, 26-Jan-2007.)
(φAB)    &   (φ𝐶 = A)    &   (φ𝐷 = B)       (φ𝐶𝐷)

Theoremsyl5eqss 2962 B chained subclass and equality deduction. (Contributed by NM, 25-Apr-2004.)
A = B    &   (φB𝐶)       (φA𝐶)

Theoremsyl5eqssr 2963 B chained subclass and equality deduction. (Contributed by NM, 25-Apr-2004.)
B = A    &   (φB𝐶)       (φA𝐶)

Theoremsyl6sseq 2964 A chained subclass and equality deduction. (Contributed by NM, 25-Apr-2004.)
(φAB)    &   B = 𝐶       (φA𝐶)

Theoremsyl6sseqr 2965 A chained subclass and equality deduction. (Contributed by NM, 25-Apr-2004.)
(φAB)    &   𝐶 = B       (φA𝐶)

Theoremsyl5sseq 2966 Subclass transitivity deduction. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.)
BA    &   (φA = 𝐶)       (φB𝐶)

Theoremsyl5sseqr 2967 Subclass transitivity deduction. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.)
BA    &   (φ𝐶 = A)       (φB𝐶)

Theoremsyl6eqss 2968 A chained subclass and equality deduction. (Contributed by Mario Carneiro, 2-Jan-2017.)
(φA = B)    &   B𝐶       (φA𝐶)

Theoremsyl6eqssr 2969 A chained subclass and equality deduction. (Contributed by Mario Carneiro, 2-Jan-2017.)
(φB = A)    &   B𝐶       (φA𝐶)

Theoremeqimss 2970 Equality implies the subclass relation. (Contributed by NM, 5-Aug-1993.) (Proof shortened by Andrew Salmon, 21-Jun-2011.)
(A = BAB)

Theoremeqimss2 2971 Equality implies the subclass relation. (Contributed by NM, 23-Nov-2003.)
(B = AAB)

Theoremeqimssi 2972 Infer subclass relationship from equality. (Contributed by NM, 6-Jan-2007.)
A = B       AB

Theoremeqimss2i 2973 Infer subclass relationship from equality. (Contributed by NM, 7-Jan-2007.)
A = B       BA

Theoremnssne1 2974 Two classes are different if they don't include the same class. (Contributed by NM, 23-Apr-2015.)
((AB ¬ A𝐶) → B𝐶)

Theoremnssne2 2975 Two classes are different if they are not subclasses of the same class. (Contributed by NM, 23-Apr-2015.)
((A𝐶 ¬ B𝐶) → AB)

Theoremnssr 2976* Negation of subclass relationship. One direction of Exercise 13 of [TakeutiZaring] p. 18. (Contributed by Jim Kingdon, 15-Jul-2018.)
(x(x A ¬ x B) → ¬ AB)

Theoremssralv 2977* Quantification restricted to a subclass. (Contributed by NM, 11-Mar-2006.)
(AB → (x B φx A φ))

Theoremssrexv 2978* Existential quantification restricted to a subclass. (Contributed by NM, 11-Jan-2007.)
(AB → (x A φx B φ))

Theoremralss 2979* Restricted universal quantification on a subset in terms of superset. (Contributed by Stefan O'Rear, 3-Apr-2015.)
(AB → (x A φx B (x Aφ)))

Theoremrexss 2980* Restricted existential quantification on a subset in terms of superset. (Contributed by Stefan O'Rear, 3-Apr-2015.)
(AB → (x A φx B (x A φ)))

Theoremss2ab 2981 Class abstractions in a subclass relationship. (Contributed by NM, 3-Jul-1994.)
({xφ} ⊆ {xψ} ↔ x(φψ))

Theoremabss 2982* Class abstraction in a subclass relationship. (Contributed by NM, 16-Aug-2006.)
({xφ} ⊆ Ax(φx A))

Theoremssab 2983* Subclass of a class abstraction. (Contributed by NM, 16-Aug-2006.)
(A ⊆ {xφ} ↔ x(x Aφ))

Theoremssabral 2984* The relation for a subclass of a class abstraction is equivalent to restricted quantification. (Contributed by NM, 6-Sep-2006.)
(A ⊆ {xφ} ↔ x A φ)

Theoremss2abi 2985 Inference of abstraction subclass from implication. (Contributed by NM, 31-Mar-1995.)
(φψ)       {xφ} ⊆ {xψ}

Theoremss2abdv 2986* Deduction of abstraction subclass from implication. (Contributed by NM, 29-Jul-2011.)
(φ → (ψχ))       (φ → {xψ} ⊆ {xχ})

Theoremabssdv 2987* Deduction of abstraction subclass from implication. (Contributed by NM, 20-Jan-2006.)
(φ → (ψx A))       (φ → {xψ} ⊆ A)

Theoremabssi 2988* Inference of abstraction subclass from implication. (Contributed by NM, 20-Jan-2006.)
(φx A)       {xφ} ⊆ A

Theoremss2rab 2989 Restricted abstraction classes in a subclass relationship. (Contributed by NM, 30-May-1999.)
({x Aφ} ⊆ {x Aψ} ↔ x A (φψ))

Theoremrabss 2990* Restricted class abstraction in a subclass relationship. (Contributed by NM, 16-Aug-2006.)
({x Aφ} ⊆ Bx A (φx B))

Theoremssrab 2991* Subclass of a restricted class abstraction. (Contributed by NM, 16-Aug-2006.)
(B ⊆ {x Aφ} ↔ (BA x B φ))

Theoremssrabdv 2992* Subclass of a restricted class abstraction (deduction rule). (Contributed by NM, 31-Aug-2006.)
(φBA)    &   ((φ x B) → ψ)       (φB ⊆ {x Aψ})

Theoremrabssdv 2993* Subclass of a restricted class abstraction (deduction rule). (Contributed by NM, 2-Feb-2015.)
((φ x A ψ) → x B)       (φ → {x Aψ} ⊆ B)

Theoremss2rabdv 2994* Deduction of restricted abstraction subclass from implication. (Contributed by NM, 30-May-2006.)
((φ x A) → (ψχ))       (φ → {x Aψ} ⊆ {x Aχ})

Theoremss2rabi 2995 Inference of restricted abstraction subclass from implication. (Contributed by NM, 14-Oct-1999.)
(x A → (φψ))       {x Aφ} ⊆ {x Aψ}

Theoremrabss2 2996* Subclass law for restricted abstraction. (Contributed by NM, 18-Dec-2004.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
(AB → {x Aφ} ⊆ {x Bφ})

Theoremssab2 2997* Subclass relation for the restriction of a class abstraction. (Contributed by NM, 31-Mar-1995.)
{x ∣ (x A φ)} ⊆ A

Theoremssrab2 2998* Subclass relation for a restricted class. (Contributed by NM, 19-Mar-1997.)
{x Aφ} ⊆ A

Theoremssrabeq 2999* If the restricting class of a restricted class abstraction is a subset of this restricted class abstraction, it is equal to this restricted class abstraction. (Contributed by Alexander van der Vekens, 31-Dec-2017.)
(𝑉 ⊆ {x 𝑉φ} ↔ 𝑉 = {x 𝑉φ})

Theoremrabssab 3000 A restricted class is a subclass of the corresponding unrestricted class. (Contributed by Mario Carneiro, 23-Dec-2016.)
{x Aφ} ⊆ {xφ}

Page List
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 7301-7354
 Copyright terms: Public domain < Previous  Next >