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Theorem List for Metamath Proof Explorer - 3901-4000   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremelneeldif 3901 The elements of a set difference and the minuend are not equal. (Contributed by AV, 21-Oct-2023.)
((𝑋𝐴𝑌 ∈ (𝐵𝐴)) → 𝑋𝑌)
 
Theoremvelcomp 3902 Characterization of setvar elements of the complement of a class. (Contributed by Andrew Salmon, 15-Jul-2011.)
(𝑥 ∈ (V ∖ 𝐴) ↔ ¬ 𝑥𝐴)
 
Theoremelin 3903 Expansion of membership in an intersection of two classes. Theorem 12 of [Suppes] p. 25. (Contributed by NM, 29-Apr-1994.)
(𝐴 ∈ (𝐵𝐶) ↔ (𝐴𝐵𝐴𝐶))
 
2.1.12  Subclasses and subsets
 
Definitiondf-ss 3904 Define the subclass relationship. Exercise 9 of [TakeutiZaring] p. 18. For example, {1, 2} ⊆ {1, 2, 3} (ex-ss 28791). Note that 𝐴𝐴 (proved in ssid 3943). Contrast this relationship with the relationship 𝐴𝐵 (as will be defined in df-pss 3906). For a more traditional definition, but requiring a dummy variable, see dfss2 3907. Other possible definitions are given by dfss3 3909, dfss4 4192, sspss 4034, ssequn1 4114, ssequn2 4117, sseqin2 4149, and ssdif0 4297.

We prefer the label "ss" ("subset") for , despite the fact that it applies to classes. It is much more common to refer to this as the subset relation than subclass, especially since most of the time the arguments are in fact sets (and for pragmatic reasons we don't want to need to use different operations for sets). The way set.mm is set up, many things are technically classes despite morally (and provably) being sets, like 1 (cf. df-1 10879 and 1ex 10971) or ( cf. df-r 10881 and reex 10962). This has to do with the fact that there are no "set expressions": classes are expressions but there are only set variables in set.mm (cf. https://us.metamath.org/downloads/grammar-ambiguity.txt 10962). This is why we use both for subclass relations and for subset relations and call it "subset". (Contributed by NM, 27-Apr-1994.)

(𝐴𝐵 ↔ (𝐴𝐵) = 𝐴)
 
Theoremdfss 3905 Variant of subclass definition df-ss 3904. (Contributed by NM, 21-Jun-1993.)
(𝐴𝐵𝐴 = (𝐴𝐵))
 
Definitiondf-pss 3906 Define proper subclass (or strict subclass) relationship between two classes. Definition 5.9 of [TakeutiZaring] p. 17. For example, {1, 2} ⊊ {1, 2, 3} (ex-pss 28792). Note that ¬ 𝐴𝐴 (proved in pssirr 4035). Contrast this relationship with the relationship 𝐴𝐵 (as defined in df-ss 3904). Other possible definitions are given by dfpss2 4020 and dfpss3 4021. (Contributed by NM, 7-Feb-1996.)
(𝐴𝐵 ↔ (𝐴𝐵𝐴𝐵))
 
Theoremdfss2 3907* Alternate definition of the subclass relationship between two classes. Definition 5.9 of [TakeutiZaring] p. 17. (Contributed by NM, 8-Jan-2002.) Avoid ax-10 2137, ax-11 2154, ax-12 2171. (Revised by SN, 16-May-2024.)
(𝐴𝐵 ↔ ∀𝑥(𝑥𝐴𝑥𝐵))
 
Theoremdfss2OLD 3908* Obsolete version of dfss2 3907 as of 16-May-2024. (Contributed by NM, 8-Jan-2002.) (Proof modification is discouraged.) (New usage is discouraged.)
(𝐴𝐵 ↔ ∀𝑥(𝑥𝐴𝑥𝐵))
 
Theoremdfss3 3909* Alternate definition of subclass relationship. (Contributed by NM, 14-Oct-1999.)
(𝐴𝐵 ↔ ∀𝑥𝐴 𝑥𝐵)
 
Theoremdfss6 3910* Alternate definition of subclass relationship. (Contributed by RP, 16-Apr-2020.)
(𝐴𝐵 ↔ ¬ ∃𝑥(𝑥𝐴 ∧ ¬ 𝑥𝐵))
 
Theoremdfss2f 3911 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.) Avoid ax-13 2372. (Revised by Gino Giotto, 19-May-2023.)
𝑥𝐴    &   𝑥𝐵       (𝐴𝐵 ↔ ∀𝑥(𝑥𝐴𝑥𝐵))
 
Theoremdfss3f 3912 Equivalence for subclass relation, using bound-variable hypotheses instead of distinct variable conditions. (Contributed by NM, 20-Mar-2004.)
𝑥𝐴    &   𝑥𝐵       (𝐴𝐵 ↔ ∀𝑥𝐴 𝑥𝐵)
 
Theoremnfss 3913 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 3914 Membership relationships follow from a subclass relationship. (Contributed by NM, 5-Aug-1993.) Avoid ax-12 2171. (Revised by SN, 27-May-2024.)
(𝐴𝐵 → (𝐶𝐴𝐶𝐵))
 
TheoremsselOLD 3915 Obsolete version of ssel 3914 as of 27-May-2024. (Contributed by NM, 5-Aug-1993.) (Proof modification is discouraged.) (New usage is discouraged.)
(𝐴𝐵 → (𝐶𝐴𝐶𝐵))
 
Theoremssel2 3916 Membership relationships follow from a subclass relationship. (Contributed by NM, 7-Jun-2004.)
((𝐴𝐵𝐶𝐴) → 𝐶𝐵)
 
Theoremsseli 3917 Membership implication from subclass relationship. (Contributed by NM, 5-Aug-1993.)
𝐴𝐵       (𝐶𝐴𝐶𝐵)
 
Theoremsselii 3918 Membership inference from subclass relationship. (Contributed by NM, 31-May-1999.)
𝐴𝐵    &   𝐶𝐴       𝐶𝐵
 
Theoremsselid 3919 Membership inference from subclass relationship. (Contributed by NM, 25-Jun-2014.)
𝐴𝐵    &   (𝜑𝐶𝐴)       (𝜑𝐶𝐵)
 
Theoremsseld 3920 Membership deduction from subclass relationship. (Contributed by NM, 15-Nov-1995.)
(𝜑𝐴𝐵)       (𝜑 → (𝐶𝐴𝐶𝐵))
 
Theoremsselda 3921 Membership deduction from subclass relationship. (Contributed by NM, 26-Jun-2014.)
(𝜑𝐴𝐵)       ((𝜑𝐶𝐴) → 𝐶𝐵)
 
Theoremsseldd 3922 Membership inference from subclass relationship. (Contributed by NM, 14-Dec-2004.)
(𝜑𝐴𝐵)    &   (𝜑𝐶𝐴)       (𝜑𝐶𝐵)
 
Theoremssneld 3923 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 → (¬ 𝐶𝐵 → ¬ 𝐶𝐴))
 
Theoremssneldd 3924 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 → ¬ 𝐶𝐵)       (𝜑 → ¬ 𝐶𝐴)
 
Theoremssriv 3925* Inference based on subclass definition. (Contributed by NM, 21-Jun-1993.)
(𝑥𝐴𝑥𝐵)       𝐴𝐵
 
Theoremssrd 3926 Deduction based on subclass definition. (Contributed by Thierry Arnoux, 8-Mar-2017.)
𝑥𝜑    &   𝑥𝐴    &   𝑥𝐵    &   (𝜑 → (𝑥𝐴𝑥𝐵))       (𝜑𝐴𝐵)
 
Theoremssrdv 3927* Deduction based on subclass definition. (Contributed by NM, 15-Nov-1995.)
(𝜑 → (𝑥𝐴𝑥𝐵))       (𝜑𝐴𝐵)
 
Theoremsstr2 3928 Transitivity of subclass relationship. Exercise 5 of [TakeutiZaring] p. 17. (Contributed by NM, 24-Jun-1993.) (Proof shortened by Andrew Salmon, 14-Jun-2011.)
(𝐴𝐵 → (𝐵𝐶𝐴𝐶))
 
Theoremsstr 3929 Transitivity of subclass relationship. Theorem 6 of [Suppes] p. 23. (Contributed by NM, 5-Sep-2003.)
((𝐴𝐵𝐵𝐶) → 𝐴𝐶)
 
Theoremsstri 3930 Subclass transitivity inference. (Contributed by NM, 5-May-2000.)
𝐴𝐵    &   𝐵𝐶       𝐴𝐶
 
Theoremsstrd 3931 Subclass transitivity deduction. (Contributed by NM, 2-Jun-2004.)
(𝜑𝐴𝐵)    &   (𝜑𝐵𝐶)       (𝜑𝐴𝐶)
 
Theoremsstrid 3932 Subclass transitivity deduction. (Contributed by NM, 6-Feb-2014.)
𝐴𝐵    &   (𝜑𝐵𝐶)       (𝜑𝐴𝐶)
 
Theoremsstrdi 3933 Subclass transitivity deduction. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.)
(𝜑𝐴𝐵)    &   𝐵𝐶       (𝜑𝐴𝐶)
 
Theoremsylan9ss 3934 A subclass transitivity deduction. (Contributed by NM, 27-Sep-2004.) (Proof shortened by Andrew Salmon, 14-Jun-2011.)
(𝜑𝐴𝐵)    &   (𝜓𝐵𝐶)       ((𝜑𝜓) → 𝐴𝐶)
 
Theoremsylan9ssr 3935 A subclass transitivity deduction. (Contributed by NM, 27-Sep-2004.)
(𝜑𝐴𝐵)    &   (𝜓𝐵𝐶)       ((𝜓𝜑) → 𝐴𝐶)
 
Theoremeqss 3936 The subclass relationship is antisymmetric. Compare Theorem 4 of [Suppes] p. 22. (Contributed by NM, 21-May-1993.)
(𝐴 = 𝐵 ↔ (𝐴𝐵𝐵𝐴))
 
Theoremeqssi 3937 Infer equality from two subclass relationships. Compare Theorem 4 of [Suppes] p. 22. (Contributed by NM, 9-Sep-1993.)
𝐴𝐵    &   𝐵𝐴       𝐴 = 𝐵
 
Theoremeqssd 3938 Equality deduction from two subclass relationships. Compare Theorem 4 of [Suppes] p. 22. (Contributed by NM, 27-Jun-2004.)
(𝜑𝐴𝐵)    &   (𝜑𝐵𝐴)       (𝜑𝐴 = 𝐵)
 
Theoremsssseq 3939 If a class is a subclass of another class, then the classes are equal if and only if the other class is a subclass of the first class. (Contributed by AV, 23-Dec-2020.)
(𝐵𝐴 → (𝐴𝐵𝐴 = 𝐵))
 
Theoremeqrd 3940 Deduce equality of classes from equivalence of membership. (Contributed by Thierry Arnoux, 21-Mar-2017.) (Proof shortened by BJ, 1-Dec-2021.)
𝑥𝜑    &   𝑥𝐴    &   𝑥𝐵    &   (𝜑 → (𝑥𝐴𝑥𝐵))       (𝜑𝐴 = 𝐵)
 
Theoremeqri 3941 Infer equality of classes from equivalence of membership. (Contributed by Thierry Arnoux, 7-Oct-2017.)
𝑥𝐴    &   𝑥𝐵    &   (𝑥𝐴𝑥𝐵)       𝐴 = 𝐵
 
Theoremeqelssd 3942* Equality deduction from subclass relationship and membership. (Contributed by AV, 21-Aug-2022.)
(𝜑𝐴𝐵)    &   ((𝜑𝑥𝐵) → 𝑥𝐴)       (𝜑𝐴 = 𝐵)
 
Theoremssid 3943 Any class is a subclass of itself. Exercise 10 of [TakeutiZaring] p. 18. (Contributed by NM, 21-Jun-1993.) (Proof shortened by Andrew Salmon, 14-Jun-2011.)
𝐴𝐴
 
Theoremssidd 3944 Weakening of ssid 3943. (Contributed by BJ, 1-Sep-2022.)
(𝜑𝐴𝐴)
 
Theoremssv 3945 Any class is a subclass of the universal class. (Contributed by NM, 31-Oct-1995.)
𝐴 ⊆ V
 
Theoremsseq1 3946 Equality theorem for subclasses. (Contributed by NM, 24-Jun-1993.) (Proof shortened by Andrew Salmon, 21-Jun-2011.)
(𝐴 = 𝐵 → (𝐴𝐶𝐵𝐶))
 
Theoremsseq2 3947 Equality theorem for the subclass relationship. (Contributed by NM, 25-Jun-1998.)
(𝐴 = 𝐵 → (𝐶𝐴𝐶𝐵))
 
Theoremsseq12 3948 Equality theorem for the subclass relationship. (Contributed by NM, 31-May-1999.)
((𝐴 = 𝐵𝐶 = 𝐷) → (𝐴𝐶𝐵𝐷))
 
Theoremsseq1i 3949 An equality inference for the subclass relationship. (Contributed by NM, 18-Aug-1993.)
𝐴 = 𝐵       (𝐴𝐶𝐵𝐶)
 
Theoremsseq2i 3950 An equality inference for the subclass relationship. (Contributed by NM, 30-Aug-1993.)
𝐴 = 𝐵       (𝐶𝐴𝐶𝐵)
 
Theoremsseq12i 3951 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 3952 An equality deduction for the subclass relationship. (Contributed by NM, 14-Aug-1994.)
(𝜑𝐴 = 𝐵)       (𝜑 → (𝐴𝐶𝐵𝐶))
 
Theoremsseq2d 3953 An equality deduction for the subclass relationship. (Contributed by NM, 14-Aug-1994.)
(𝜑𝐴 = 𝐵)       (𝜑 → (𝐶𝐴𝐶𝐵))
 
Theoremsseq12d 3954 An equality deduction for the subclass relationship. (Contributed by NM, 31-May-1999.)
(𝜑𝐴 = 𝐵)    &   (𝜑𝐶 = 𝐷)       (𝜑 → (𝐴𝐶𝐵𝐷))
 
Theoremeqsstri 3955 Substitution of equality into a subclass relationship. (Contributed by NM, 16-Jul-1995.)
𝐴 = 𝐵    &   𝐵𝐶       𝐴𝐶
 
Theoremeqsstrri 3956 Substitution of equality into a subclass relationship. (Contributed by NM, 19-Oct-1999.)
𝐵 = 𝐴    &   𝐵𝐶       𝐴𝐶
 
Theoremsseqtri 3957 Substitution of equality into a subclass relationship. (Contributed by NM, 28-Jul-1995.)
𝐴𝐵    &   𝐵 = 𝐶       𝐴𝐶
 
Theoremsseqtrri 3958 Substitution of equality into a subclass relationship. (Contributed by NM, 4-Apr-1995.)
𝐴𝐵    &   𝐶 = 𝐵       𝐴𝐶
 
Theoremeqsstrd 3959 Substitution of equality into a subclass relationship. (Contributed by NM, 25-Apr-2004.)
(𝜑𝐴 = 𝐵)    &   (𝜑𝐵𝐶)       (𝜑𝐴𝐶)
 
Theoremeqsstrrd 3960 Substitution of equality into a subclass relationship. (Contributed by NM, 25-Apr-2004.)
(𝜑𝐵 = 𝐴)    &   (𝜑𝐵𝐶)       (𝜑𝐴𝐶)
 
Theoremsseqtrd 3961 Substitution of equality into a subclass relationship. (Contributed by NM, 25-Apr-2004.)
(𝜑𝐴𝐵)    &   (𝜑𝐵 = 𝐶)       (𝜑𝐴𝐶)
 
Theoremsseqtrrd 3962 Substitution of equality into a subclass relationship. (Contributed by NM, 25-Apr-2004.)
(𝜑𝐴𝐵)    &   (𝜑𝐶 = 𝐵)       (𝜑𝐴𝐶)
 
Theorem3sstr3i 3963 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 3964 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 3965 Substitution of equality into both sides of a subclass relationship. (Contributed by NM, 1-Oct-2000.)
(𝜑𝐴𝐵)    &   𝐴 = 𝐶    &   𝐵 = 𝐷       (𝜑𝐶𝐷)
 
Theorem3sstr4g 3966 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 3967 Substitution of equality into both sides of a subclass relationship. (Contributed by NM, 1-Oct-2000.)
(𝜑𝐴𝐵)    &   (𝜑𝐴 = 𝐶)    &   (𝜑𝐵 = 𝐷)       (𝜑𝐶𝐷)
 
Theorem3sstr4d 3968 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)
 
Theoremeqsstrid 3969 A chained subclass and equality deduction. (Contributed by NM, 25-Apr-2004.)
𝐴 = 𝐵    &   (𝜑𝐵𝐶)       (𝜑𝐴𝐶)
 
Theoremeqsstrrid 3970 A chained subclass and equality deduction. (Contributed by NM, 25-Apr-2004.)
𝐵 = 𝐴    &   (𝜑𝐵𝐶)       (𝜑𝐴𝐶)
 
Theoremsseqtrdi 3971 A chained subclass and equality deduction. (Contributed by NM, 25-Apr-2004.)
(𝜑𝐴𝐵)    &   𝐵 = 𝐶       (𝜑𝐴𝐶)
 
Theoremsseqtrrdi 3972 A chained subclass and equality deduction. (Contributed by NM, 25-Apr-2004.)
(𝜑𝐴𝐵)    &   𝐶 = 𝐵       (𝜑𝐴𝐶)
 
Theoremsseqtrid 3973 Subclass transitivity deduction. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.)
𝐵𝐴    &   (𝜑𝐴 = 𝐶)       (𝜑𝐵𝐶)
 
Theoremsseqtrrid 3974 Subclass transitivity deduction. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.)
𝐵𝐴    &   (𝜑𝐶 = 𝐴)       (𝜑𝐵𝐶)
 
Theoremeqsstrdi 3975 A chained subclass and equality deduction. (Contributed by Mario Carneiro, 2-Jan-2017.)
(𝜑𝐴 = 𝐵)    &   𝐵𝐶       (𝜑𝐴𝐶)
 
Theoremeqsstrrdi 3976 A chained subclass and equality deduction. (Contributed by Mario Carneiro, 2-Jan-2017.)
(𝜑𝐵 = 𝐴)    &   𝐵𝐶       (𝜑𝐴𝐶)
 
Theoremeqimss 3977 Equality implies inclusion. (Contributed by NM, 21-Jun-1993.) (Proof shortened by Andrew Salmon, 21-Jun-2011.)
(𝐴 = 𝐵𝐴𝐵)
 
Theoremeqimss2 3978 Equality implies inclusion. (Contributed by NM, 23-Nov-2003.)
(𝐵 = 𝐴𝐴𝐵)
 
Theoremeqimssi 3979 Infer subclass relationship from equality. (Contributed by NM, 6-Jan-2007.)
𝐴 = 𝐵       𝐴𝐵
 
Theoremeqimss2i 3980 Infer subclass relationship from equality. (Contributed by NM, 7-Jan-2007.)
𝐴 = 𝐵       𝐵𝐴
 
Theoremnssne1 3981 Two classes are different if they don't include the same class. (Contributed by NM, 23-Apr-2015.)
((𝐴𝐵 ∧ ¬ 𝐴𝐶) → 𝐵𝐶)
 
Theoremnssne2 3982 Two classes are different if they are not subclasses of the same class. (Contributed by NM, 23-Apr-2015.)
((𝐴𝐶 ∧ ¬ 𝐵𝐶) → 𝐴𝐵)
 
Theoremnss 3983* Negation of subclass relationship. Exercise 13 of [TakeutiZaring] p. 18. (Contributed by NM, 25-Feb-1996.) (Proof shortened by Andrew Salmon, 21-Jun-2011.)
𝐴𝐵 ↔ ∃𝑥(𝑥𝐴 ∧ ¬ 𝑥𝐵))
 
Theoremnelss 3984 Demonstrate by witnesses that two classes lack a subclass relation. (Contributed by Stefan O'Rear, 5-Feb-2015.)
((𝐴𝐵 ∧ ¬ 𝐴𝐶) → ¬ 𝐵𝐶)
 
Theoremssrexf 3985 Restricted existential quantification follows from a subclass relationship. (Contributed by Glauco Siliprandi, 20-Apr-2017.)
𝑥𝐴    &   𝑥𝐵       (𝐴𝐵 → (∃𝑥𝐴 𝜑 → ∃𝑥𝐵 𝜑))
 
Theoremssrmof 3986 "At most one" existential quantification restricted to a subclass. (Contributed by Thierry Arnoux, 8-Oct-2017.)
𝑥𝐴    &   𝑥𝐵       (𝐴𝐵 → (∃*𝑥𝐵 𝜑 → ∃*𝑥𝐴 𝜑))
 
Theoremssralv 3987* Quantification restricted to a subclass. (Contributed by NM, 11-Mar-2006.)
(𝐴𝐵 → (∀𝑥𝐵 𝜑 → ∀𝑥𝐴 𝜑))
 
Theoremssrexv 3988* Existential quantification restricted to a subclass. (Contributed by NM, 11-Jan-2007.)
(𝐴𝐵 → (∃𝑥𝐴 𝜑 → ∃𝑥𝐵 𝜑))
 
Theoremss2ralv 3989* Two quantifications restricted to a subclass. (Contributed by AV, 11-Mar-2023.)
(𝐴𝐵 → (∀𝑥𝐵𝑦𝐵 𝜑 → ∀𝑥𝐴𝑦𝐴 𝜑))
 
Theoremss2rexv 3990* Two existential quantifications restricted to a subclass. (Contributed by AV, 11-Mar-2023.)
(𝐴𝐵 → (∃𝑥𝐴𝑦𝐴 𝜑 → ∃𝑥𝐵𝑦𝐵 𝜑))
 
Theoremralss 3991* Restricted universal quantification on a subset in terms of superset. (Contributed by Stefan O'Rear, 3-Apr-2015.)
(𝐴𝐵 → (∀𝑥𝐴 𝜑 ↔ ∀𝑥𝐵 (𝑥𝐴𝜑)))
 
Theoremrexss 3992* Restricted existential quantification on a subset in terms of superset. (Contributed by Stefan O'Rear, 3-Apr-2015.)
(𝐴𝐵 → (∃𝑥𝐴 𝜑 ↔ ∃𝑥𝐵 (𝑥𝐴𝜑)))
 
Theoremss2ab 3993 Class abstractions in a subclass relationship. (Contributed by NM, 3-Jul-1994.)
({𝑥𝜑} ⊆ {𝑥𝜓} ↔ ∀𝑥(𝜑𝜓))
 
Theoremabss 3994* Class abstraction in a subclass relationship. (Contributed by NM, 16-Aug-2006.)
({𝑥𝜑} ⊆ 𝐴 ↔ ∀𝑥(𝜑𝑥𝐴))
 
Theoremssab 3995* Subclass of a class abstraction. (Contributed by NM, 16-Aug-2006.)
(𝐴 ⊆ {𝑥𝜑} ↔ ∀𝑥(𝑥𝐴𝜑))
 
Theoremssabral 3996* The relation for a subclass of a class abstraction is equivalent to restricted quantification. (Contributed by NM, 6-Sep-2006.)
(𝐴 ⊆ {𝑥𝜑} ↔ ∀𝑥𝐴 𝜑)
 
Theoremss2abdv 3997* Deduction of abstraction subclass from implication. (Contributed by NM, 29-Jul-2011.) Avoid ax-8 2108, ax-10 2137, ax-11 2154, ax-12 2171. (Revised by Gino Giotto, 28-Jun-2024.)
(𝜑 → (𝜓𝜒))       (𝜑 → {𝑥𝜓} ⊆ {𝑥𝜒})
 
Theoremss2abdvALT 3998* Alternate proof of ss2abdv 3997. Shorter, but requiring ax-8 2108. (Contributed by Steven Nguyen, 28-Jun-2024.) (Proof modification is discouraged.) (New usage is discouraged.)
(𝜑 → (𝜓𝜒))       (𝜑 → {𝑥𝜓} ⊆ {𝑥𝜒})
 
Theoremss2abdvOLD 3999* Obsolete version of ss2abdv 3997 as of 28-Jun-2024. (Contributed by NM, 29-Jul-2011.) (Proof modification is discouraged.) (New usage is discouraged.)
(𝜑 → (𝜓𝜒))       (𝜑 → {𝑥𝜓} ⊆ {𝑥𝜒})
 
Theoremss2abi 4000 Inference of abstraction subclass from implication. (Contributed by NM, 31-Mar-1995.) Avoid ax-8 2108, ax-10 2137, ax-11 2154, ax-12 2171. (Revised by Gino Giotto, 28-Jun-2024.)
(𝜑𝜓)       {𝑥𝜑} ⊆ {𝑥𝜓}
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