P. 40 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 3901-4000   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	cbvrabcsf 3901	A more general version of cbvrab 3457 with no distinct variable restrictions. Usage of this theorem is discouraged because it depends on ax-13 2407. (Contributed by Andrew Salmon, 13-Jul-2011.) (New usage is discouraged.)
⊢ Ⅎ𝑦𝐴    &   ⊢ Ⅎ𝑥𝐵    &   ⊢ Ⅎ𝑦𝜑    &   ⊢ Ⅎ𝑥𝜓    &   ⊢ (𝑥 = 𝑦 → 𝐴 = 𝐵)    &   ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))    ⇒   ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑦 ∈ 𝐵 ∣ 𝜓}
Theorem	cbvralv2 3902*	Rule used to change the bound variable in a restricted universal quantifier with implicit substitution which also changes the quantifier domain. Usage of this theorem is discouraged because it depends on ax-13 2407. (Contributed by David Moews, 1-May-2017.) (New usage is discouraged.)
⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜒))    &   ⊢ (𝑥 = 𝑦 → 𝐴 = 𝐵)    ⇒   ⊢ (∀𝑥 ∈ 𝐴 𝜓 ↔ ∀𝑦 ∈ 𝐵 𝜒)
Theorem	cbvrexv2 3903*	Rule used to change the bound variable in a restricted existential quantifier with implicit substitution which also changes the quantifier domain. Usage of this theorem is discouraged because it depends on ax-13 2407. (Contributed by David Moews, 1-May-2017.) (New usage is discouraged.)
⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜒))    &   ⊢ (𝑥 = 𝑦 → 𝐴 = 𝐵)    ⇒   ⊢ (∃𝑥 ∈ 𝐴 𝜓 ↔ ∃𝑦 ∈ 𝐵 𝜒)
Theorem	rspc2vd 3904*	Deduction version of 2-variable restricted specialization, using implicit substitution. Notice that the class 𝐷 for the second set variable 𝑦 may depend on the first set variable 𝑥. (Contributed by AV, 29-Mar-2021.)
⊢ (𝑥 = 𝐴 → (𝜃 ↔ 𝜒))    &   ⊢ (𝑦 = 𝐵 → (𝜒 ↔ 𝜓))    &   ⊢ (𝜑 → 𝐴 ∈ 𝐶)    &   ⊢ ((𝜑 ∧ 𝑥 = 𝐴) → 𝐷 = 𝐸)    &   ⊢ (𝜑 → 𝐵 ∈ 𝐸)    ⇒   ⊢ (𝜑 → (∀𝑥 ∈ 𝐶 ∀𝑦 ∈ 𝐷 𝜃 → 𝜓))
2.1.11  Define basic set operations and relations
Syntax	cdif 3905	Extend class notation to include class difference (read: "𝐴 minus 𝐵").
class (𝐴 ∖ 𝐵)
Syntax	cun 3906	Extend class notation to include union of two classes (read: "𝐴 union 𝐵").
class (𝐴 ∪ 𝐵)
Syntax	cin 3907	Extend class notation to include the intersection of two classes (read: "𝐴 intersect 𝐵").
class (𝐴 ∩ 𝐵)
Syntax	wss 3908	Extend wff notation to include the subclass relation. This is read "𝐴 is a subclass of 𝐵 " or "𝐵 includes 𝐴". When 𝐴 exists as a set, it is also read "𝐴 is a subset of 𝐵".
wff 𝐴 ⊆ 𝐵
Syntax	wpss 3909	Extend wff notation with proper subclass relation.
wff 𝐴 ⊊ 𝐵
Theorem	difjust 3910*	Soundness justification theorem for df-dif 3911. (Contributed by Rodolfo Medina, 27-Apr-2010.) (Proof shortened by Andrew Salmon, 9-Jul-2011.)
⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵)} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ ¬ 𝑦 ∈ 𝐵)}
Definition	df-dif 3911*	Define class difference, also called relative complement. Definition 5.12 of [TakeutiZaring] p. 20. For example, ({1, 3} ∖ {1, 8}) = {3} (ex-dif 30789). Contrast this operation with union (𝐴 ∪ 𝐵) (df-un 3913) and intersection (𝐴 ∩ 𝐵) (df-in 3915). Several notations are used in the literature; we chose the ∖ convention used in Definition 5.3 of [Eisenberg] p. 67 instead of the more common minus sign to reserve the latter for later use in, e.g., arithmetic. We will use the terminology "𝐴 excludes 𝐵 " to mean 𝐴 ∖ 𝐵. We will use "𝐵 is removed from 𝐴 " to mean 𝐴 ∖ {𝐵} i.e. the removal of an element or equivalently the exclusion of a singleton. (Contributed by NM, 29-Apr-1994.)
⊢ (𝐴 ∖ 𝐵) = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵)}
Theorem	unjust 3912*	Soundness justification theorem for df-un 3913. (Contributed by Rodolfo Medina, 28-Apr-2010.) (Proof shortened by Andrew Salmon, 9-Jul-2011.)
⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵)} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝐵)}
Definition	df-un 3913*	Define the union of two classes. Definition 5.6 of [TakeutiZaring] p. 16. For example, ({1, 3} ∪ {1, 8}) = {1, 3, 8} (ex-un 30790). Contrast this operation with difference (𝐴 ∖ 𝐵) (df-dif 3911) and intersection (𝐴 ∩ 𝐵) (df-in 3915). For an alternate definition in terms of class difference, requiring no dummy variables, see dfun2 4226. For union defined in terms of intersection, see dfun3 4232. (Contributed by NM, 23-Aug-1993.)
⊢ (𝐴 ∪ 𝐵) = {𝑥 ∣ (𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵)}
Theorem	injust 3914*	Soundness justification theorem for df-in 3915. (Contributed by Rodolfo Medina, 28-Apr-2010.) (Proof shortened by Andrew Salmon, 9-Jul-2011.)
⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {𝑦 ∣ (𝑦 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)}
Definition	df-in 3915*	Define the intersection of two classes. Definition 5.6 of [TakeutiZaring] p. 16. For example, ({1, 3} ∩ {1, 8}) = {1} (ex-in 30791). Contrast this operation with union (𝐴 ∪ 𝐵) (df-un 3913) and difference (𝐴 ∖ 𝐵) (df-dif 3911). For alternate definitions in terms of class difference, requiring no dummy variables, see dfin2 4227 and dfin4 4234. For intersection defined in terms of union, see dfin3 4233. (Contributed by NM, 29-Apr-1994.)
⊢ (𝐴 ∩ 𝐵) = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)}
Theorem	dfin5 3916*	Alternate definition for the intersection of two classes. (Contributed by NM, 6-Jul-2005.)
⊢ (𝐴 ∩ 𝐵) = {𝑥 ∈ 𝐴 ∣ 𝑥 ∈ 𝐵}
Theorem	dfdif2 3917*	Alternate definition of class difference. (Contributed by NM, 25-Mar-2004.)
⊢ (𝐴 ∖ 𝐵) = {𝑥 ∈ 𝐴 ∣ ¬ 𝑥 ∈ 𝐵}
Theorem	eldif 3918	Expansion of membership in a class difference. (Contributed by NM, 29-Apr-1994.)
⊢ (𝐴 ∈ (𝐵 ∖ 𝐶) ↔ (𝐴 ∈ 𝐵 ∧ ¬ 𝐴 ∈ 𝐶))
Theorem	eldifd 3919	If a class is in one class and not another, it is also in their difference. One-way deduction form of eldif 3918. (Contributed by David Moews, 1-May-2017.)
⊢ (𝜑 → 𝐴 ∈ 𝐵)    &   ⊢ (𝜑 → ¬ 𝐴 ∈ 𝐶)    ⇒   ⊢ (𝜑 → 𝐴 ∈ (𝐵 ∖ 𝐶))
Theorem	eldifad 3920	If a class is in the difference of two classes, it is also in the minuend. One-way deduction form of eldif 3918. (Contributed by David Moews, 1-May-2017.)
⊢ (𝜑 → 𝐴 ∈ (𝐵 ∖ 𝐶))    ⇒   ⊢ (𝜑 → 𝐴 ∈ 𝐵)
Theorem	eldifbd 3921	If a class is in the difference of two classes, it is not in the subtrahend. One-way deduction form of eldif 3918. (Contributed by David Moews, 1-May-2017.)
⊢ (𝜑 → 𝐴 ∈ (𝐵 ∖ 𝐶))    ⇒   ⊢ (𝜑 → ¬ 𝐴 ∈ 𝐶)
Theorem	elneeldif 3922	The elements of a set difference and the minuend are not equal. (Contributed by AV, 21-Oct-2023.)
⊢ ((𝑋 ∈ 𝐴 ∧ 𝑌 ∈ (𝐵 ∖ 𝐴)) → 𝑋 ≠ 𝑌)
Theorem	velcomp 3923	Characterization of setvar elements of the complement of a class. (Contributed by Andrew Salmon, 15-Jul-2011.)
⊢ (𝑥 ∈ (V ∖ 𝐴) ↔ ¬ 𝑥 ∈ 𝐴)
Theorem	elin 3924	Expansion of membership in an intersection of two classes. Theorem 12 of [Suppes] p. 25. (Contributed by NM, 29-Apr-1994.)
⊢ (𝐴 ∈ (𝐵 ∩ 𝐶) ↔ (𝐴 ∈ 𝐵 ∧ 𝐴 ∈ 𝐶))
2.1.12  Subclasses and subsets
Definition	df-ss 3925*	Define the subclass relationship. Definition 5.9 of [TakeutiZaring] p. 17. For example, {1, 2} ⊆ {1, 2, 3} (ex-ss 30793). Note that 𝐴 ⊆ 𝐴 (proved in ssid 3962). Contrast this relationship with the relationship 𝐴 ⊊ 𝐵 (as will be defined in df-pss 3928). For an alternative definition, not requiring a dummy variable, see dfss2 3926. Other possible definitions are given by dfss3 3929, dfss4 4225, sspss 4059, ssequn1 4142, ssequn2 4145, sseqin2 4179, and ssdif0 4324. 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 11118 and 1ex 11213) or ℝ ( cf. df-r 11120 and reex 11201). 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 11201). This is why we use ⊆ both for subclass relations and for subset relations and call it "subset". (Contributed by NM, 8-Jan-2002.) Revised from the original definition dfss2 3926. (Revised by GG, 15-May-2025.)
⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
Theorem	dfss2 3926	Alternate definition of the subclass relationship between two classes. Exercise 9 of [TakeutiZaring] p. 18. This was the original definition before df-ss 3925. (Contributed by NM, 27-Apr-1994.) Revise df-ss 3925. (Revised by GG, 15-May-2025.)
⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ∩ 𝐵) = 𝐴)
Theorem	dfss 3927	Variant of subclass definition dfss2 3926. (Contributed by NM, 21-Jun-1993.)
⊢ (𝐴 ⊆ 𝐵 ↔ 𝐴 = (𝐴 ∩ 𝐵))
Definition	df-pss 3928	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 30794). Note that ¬ 𝐴 ⊊ 𝐴 (proved in pssirr 4060). Contrast this relationship with the relationship 𝐴 ⊆ 𝐵 (as defined in df-ss 3925). Other possible definitions are given by dfpss2 4045 and dfpss3 4046. (Contributed by NM, 7-Feb-1996.)
⊢ (𝐴 ⊊ 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐴 ≠ 𝐵))
Theorem	dfss3 3929*	Alternate definition of subclass relationship. (Contributed by NM, 14-Oct-1999.)
⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥 ∈ 𝐴 𝑥 ∈ 𝐵)
Theorem	dfss6 3930*	Alternate definition of subclass relationship. (Contributed by RP, 16-Apr-2020.)
⊢ (𝐴 ⊆ 𝐵 ↔ ¬ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
Theorem	dfssf 3931	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 2407. (Revised by GG, 19-May-2023.)
⊢ Ⅎ𝑥𝐴    &   ⊢ Ⅎ𝑥𝐵    ⇒   ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
Theorem	dfss3f 3932	Equivalence for subclass relation, using bound-variable hypotheses instead of distinct variable conditions. (Contributed by NM, 20-Mar-2004.)
⊢ Ⅎ𝑥𝐴    &   ⊢ Ⅎ𝑥𝐵    ⇒   ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥 ∈ 𝐴 𝑥 ∈ 𝐵)
Theorem	nfss 3933	If 𝑥 is not free in 𝐴 and 𝐵, it is not free in 𝐴 ⊆ 𝐵. (Contributed by NM, 27-Dec-1996.)
⊢ Ⅎ𝑥𝐴    &   ⊢ Ⅎ𝑥𝐵    ⇒   ⊢ Ⅎ𝑥 𝐴 ⊆ 𝐵
Theorem	ssel 3934	Membership relationships follow from a subclass relationship. (Contributed by NM, 5-Aug-1993.) Avoid ax-12 2216. (Revised by SN, 27-May-2024.)
⊢ (𝐴 ⊆ 𝐵 → (𝐶 ∈ 𝐴 → 𝐶 ∈ 𝐵))
Theorem	ssel2 3935	Membership relationships follow from a subclass relationship. (Contributed by NM, 7-Jun-2004.)
⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐶 ∈ 𝐴) → 𝐶 ∈ 𝐵)
Theorem	sseli 3936	Membership implication from subclass relationship. (Contributed by NM, 5-Aug-1993.)
⊢ 𝐴 ⊆ 𝐵    ⇒   ⊢ (𝐶 ∈ 𝐴 → 𝐶 ∈ 𝐵)
Theorem	sselii 3937	Membership inference from subclass relationship. (Contributed by NM, 31-May-1999.)
⊢ 𝐴 ⊆ 𝐵    &   ⊢ 𝐶 ∈ 𝐴    ⇒   ⊢ 𝐶 ∈ 𝐵
Theorem	sselid 3938	Membership inference from subclass relationship. (Contributed by NM, 25-Jun-2014.)
⊢ 𝐴 ⊆ 𝐵    &   ⊢ (𝜑 → 𝐶 ∈ 𝐴)    ⇒   ⊢ (𝜑 → 𝐶 ∈ 𝐵)
Theorem	sseld 3939	Membership deduction from subclass relationship. (Contributed by NM, 15-Nov-1995.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    ⇒   ⊢ (𝜑 → (𝐶 ∈ 𝐴 → 𝐶 ∈ 𝐵))
Theorem	sselda 3940	Membership deduction from subclass relationship. (Contributed by NM, 26-Jun-2014.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    ⇒   ⊢ ((𝜑 ∧ 𝐶 ∈ 𝐴) → 𝐶 ∈ 𝐵)
Theorem	sseldd 3941	Membership inference from subclass relationship. (Contributed by NM, 14-Dec-2004.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    &   ⊢ (𝜑 → 𝐶 ∈ 𝐴)    ⇒   ⊢ (𝜑 → 𝐶 ∈ 𝐵)
Theorem	ssneld 3942	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.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    ⇒   ⊢ (𝜑 → (¬ 𝐶 ∈ 𝐵 → ¬ 𝐶 ∈ 𝐴))
Theorem	ssneldd 3943	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.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    &   ⊢ (𝜑 → ¬ 𝐶 ∈ 𝐵)    ⇒   ⊢ (𝜑 → ¬ 𝐶 ∈ 𝐴)
Theorem	ssriv 3944*	Inference based on subclass definition. (Contributed by NM, 21-Jun-1993.)
⊢ (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵)    ⇒   ⊢ 𝐴 ⊆ 𝐵
Theorem	ssrd 3945	Deduction based on subclass definition. (Contributed by Thierry Arnoux, 8-Mar-2017.)
⊢ Ⅎ𝑥𝜑    &   ⊢ Ⅎ𝑥𝐴    &   ⊢ Ⅎ𝑥𝐵    &   ⊢ (𝜑 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))    ⇒   ⊢ (𝜑 → 𝐴 ⊆ 𝐵)
Theorem	ssrdv 3946*	Deduction based on subclass definition. (Contributed by NM, 15-Nov-1995.)
⊢ (𝜑 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))    ⇒   ⊢ (𝜑 → 𝐴 ⊆ 𝐵)
Theorem	sstr2 3947	Transitivity of subclass relationship. Exercise 5 of [TakeutiZaring] p. 17. (Contributed by NM, 24-Jun-1993.) (Proof shortened by Andrew Salmon, 14-Jun-2011.) Avoid axioms. (Revised by GG, 19-May-2025.)
⊢ (𝐴 ⊆ 𝐵 → (𝐵 ⊆ 𝐶 → 𝐴 ⊆ 𝐶))
Theorem	sstr 3948	Transitivity of subclass relationship. Theorem 6 of [Suppes] p. 23. (Contributed by NM, 5-Sep-2003.)
⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐶) → 𝐴 ⊆ 𝐶)
Theorem	sstri 3949	Subclass transitivity inference. (Contributed by NM, 5-May-2000.)
⊢ 𝐴 ⊆ 𝐵    &   ⊢ 𝐵 ⊆ 𝐶    ⇒   ⊢ 𝐴 ⊆ 𝐶
Theorem	sstrd 3950	Subclass transitivity deduction. (Contributed by NM, 2-Jun-2004.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    &   ⊢ (𝜑 → 𝐵 ⊆ 𝐶)    ⇒   ⊢ (𝜑 → 𝐴 ⊆ 𝐶)
Theorem	sstrid 3951	Subclass transitivity deduction. (Contributed by NM, 6-Feb-2014.)
⊢ 𝐴 ⊆ 𝐵    &   ⊢ (𝜑 → 𝐵 ⊆ 𝐶)    ⇒   ⊢ (𝜑 → 𝐴 ⊆ 𝐶)
Theorem	sstrdi 3952	Subclass transitivity deduction. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    &   ⊢ 𝐵 ⊆ 𝐶    ⇒   ⊢ (𝜑 → 𝐴 ⊆ 𝐶)
Theorem	sylan9ss 3953	A subclass transitivity deduction. (Contributed by NM, 27-Sep-2004.) (Proof shortened by Andrew Salmon, 14-Jun-2011.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    &   ⊢ (𝜓 → 𝐵 ⊆ 𝐶)    ⇒   ⊢ ((𝜑 ∧ 𝜓) → 𝐴 ⊆ 𝐶)
Theorem	sylan9ssr 3954	A subclass transitivity deduction. (Contributed by NM, 27-Sep-2004.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    &   ⊢ (𝜓 → 𝐵 ⊆ 𝐶)    ⇒   ⊢ ((𝜓 ∧ 𝜑) → 𝐴 ⊆ 𝐶)
Theorem	eqss 3955	The subclass relationship is antisymmetric. Compare Theorem 4 of [Suppes] p. 22. (Contributed by NM, 21-May-1993.)
⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
Theorem	eqssi 3956	Infer equality from two subclass relationships. Compare Theorem 4 of [Suppes] p. 22. (Contributed by NM, 9-Sep-1993.)
⊢ 𝐴 ⊆ 𝐵    &   ⊢ 𝐵 ⊆ 𝐴    ⇒   ⊢ 𝐴 = 𝐵
Theorem	eqssd 3957	Equality deduction from two subclass relationships. Compare Theorem 4 of [Suppes] p. 22. (Contributed by NM, 27-Jun-2004.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    &   ⊢ (𝜑 → 𝐵 ⊆ 𝐴)    ⇒   ⊢ (𝜑 → 𝐴 = 𝐵)
Theorem	sssseq 3958	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.)
⊢ (𝐵 ⊆ 𝐴 → (𝐴 ⊆ 𝐵 ↔ 𝐴 = 𝐵))
Theorem	eqrd 3959	Deduce equality of classes from equivalence of membership. (Contributed by Thierry Arnoux, 21-Mar-2017.) (Proof shortened by BJ, 1-Dec-2021.)
⊢ Ⅎ𝑥𝜑    &   ⊢ Ⅎ𝑥𝐴    &   ⊢ Ⅎ𝑥𝐵    &   ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))    ⇒   ⊢ (𝜑 → 𝐴 = 𝐵)
Theorem	eqri 3960	Infer equality of classes from equivalence of membership. (Contributed by Thierry Arnoux, 7-Oct-2017.)
⊢ Ⅎ𝑥𝐴    &   ⊢ Ⅎ𝑥𝐵    &   ⊢ (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)    ⇒   ⊢ 𝐴 = 𝐵
Theorem	eqelssd 3961*	Equality deduction from subclass relationship and membership. (Contributed by AV, 21-Aug-2022.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐵) → 𝑥 ∈ 𝐴)    ⇒   ⊢ (𝜑 → 𝐴 = 𝐵)
Theorem	ssid 3962	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.)
⊢ 𝐴 ⊆ 𝐴
Theorem	ssidd 3963	Weakening of ssid 3962. (Contributed by BJ, 1-Sep-2022.)
⊢ (𝜑 → 𝐴 ⊆ 𝐴)
Theorem	ssv 3964	Any class is a subclass of the universal class. Dual of 0ss 4360. (Contributed by NM, 31-Oct-1995.)
⊢ 𝐴 ⊆ V
Theorem	sseq1 3965	Equality theorem for subclasses. (Contributed by NM, 24-Jun-1993.) (Proof shortened by Andrew Salmon, 21-Jun-2011.)
⊢ (𝐴 = 𝐵 → (𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐶))
Theorem	sseq2 3966	Equality theorem for the subclass relationship. (Contributed by NM, 25-Jun-1998.)
⊢ (𝐴 = 𝐵 → (𝐶 ⊆ 𝐴 ↔ 𝐶 ⊆ 𝐵))
Theorem	sseq12 3967	Equality theorem for the subclass relationship. (Contributed by NM, 31-May-1999.)
⊢ ((𝐴 = 𝐵 ∧ 𝐶 = 𝐷) → (𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐷))
Theorem	sseq1i 3968	An equality inference for the subclass relationship. (Contributed by NM, 18-Aug-1993.)
⊢ 𝐴 = 𝐵    ⇒   ⊢ (𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐶)
Theorem	sseq2i 3969	An equality inference for the subclass relationship. (Contributed by NM, 30-Aug-1993.)
⊢ 𝐴 = 𝐵    ⇒   ⊢ (𝐶 ⊆ 𝐴 ↔ 𝐶 ⊆ 𝐵)
Theorem	sseq12i 3970	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    ⇒   ⊢ (𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐷)
Theorem	sseq1d 3971	An equality deduction for the subclass relationship. (Contributed by NM, 14-Aug-1994.)
⊢ (𝜑 → 𝐴 = 𝐵)    ⇒   ⊢ (𝜑 → (𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐶))
Theorem	sseq2d 3972	An equality deduction for the subclass relationship. (Contributed by NM, 14-Aug-1994.)
⊢ (𝜑 → 𝐴 = 𝐵)    ⇒   ⊢ (𝜑 → (𝐶 ⊆ 𝐴 ↔ 𝐶 ⊆ 𝐵))
Theorem	sseq12d 3973	An equality deduction for the subclass relationship. (Contributed by NM, 31-May-1999.)
⊢ (𝜑 → 𝐴 = 𝐵)    &   ⊢ (𝜑 → 𝐶 = 𝐷)    ⇒   ⊢ (𝜑 → (𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐷))
Theorem	eqsstrd 3974	Substitution of equality into a subclass relationship. (Contributed by NM, 25-Apr-2004.)
⊢ (𝜑 → 𝐴 = 𝐵)    &   ⊢ (𝜑 → 𝐵 ⊆ 𝐶)    ⇒   ⊢ (𝜑 → 𝐴 ⊆ 𝐶)
Theorem	eqsstrrd 3975	Substitution of equality into a subclass relationship. (Contributed by NM, 25-Apr-2004.)
⊢ (𝜑 → 𝐵 = 𝐴)    &   ⊢ (𝜑 → 𝐵 ⊆ 𝐶)    ⇒   ⊢ (𝜑 → 𝐴 ⊆ 𝐶)
Theorem	sseqtrd 3976	Substitution of equality into a subclass relationship. (Contributed by NM, 25-Apr-2004.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    &   ⊢ (𝜑 → 𝐵 = 𝐶)    ⇒   ⊢ (𝜑 → 𝐴 ⊆ 𝐶)
Theorem	sseqtrrd 3977	Substitution of equality into a subclass relationship. (Contributed by NM, 25-Apr-2004.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    &   ⊢ (𝜑 → 𝐶 = 𝐵)    ⇒   ⊢ (𝜑 → 𝐴 ⊆ 𝐶)
Theorem	eqsstrid 3978	A chained subclass and equality deduction. (Contributed by NM, 25-Apr-2004.)
⊢ 𝐴 = 𝐵    &   ⊢ (𝜑 → 𝐵 ⊆ 𝐶)    ⇒   ⊢ (𝜑 → 𝐴 ⊆ 𝐶)
Theorem	eqsstrrid 3979	A chained subclass and equality deduction. (Contributed by NM, 25-Apr-2004.)
⊢ 𝐵 = 𝐴    &   ⊢ (𝜑 → 𝐵 ⊆ 𝐶)    ⇒   ⊢ (𝜑 → 𝐴 ⊆ 𝐶)
Theorem	sseqtrdi 3980	A chained subclass and equality deduction. (Contributed by NM, 25-Apr-2004.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    &   ⊢ 𝐵 = 𝐶    ⇒   ⊢ (𝜑 → 𝐴 ⊆ 𝐶)
Theorem	sseqtrrdi 3981	A chained subclass and equality deduction. (Contributed by NM, 25-Apr-2004.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    &   ⊢ 𝐶 = 𝐵    ⇒   ⊢ (𝜑 → 𝐴 ⊆ 𝐶)
Theorem	sseqtrid 3982	Subclass transitivity deduction. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.)
⊢ 𝐵 ⊆ 𝐴    &   ⊢ (𝜑 → 𝐴 = 𝐶)    ⇒   ⊢ (𝜑 → 𝐵 ⊆ 𝐶)
Theorem	sseqtrrid 3983	Subclass transitivity deduction. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.)
⊢ 𝐵 ⊆ 𝐴    &   ⊢ (𝜑 → 𝐶 = 𝐴)    ⇒   ⊢ (𝜑 → 𝐵 ⊆ 𝐶)
Theorem	eqsstrdi 3984	A chained subclass and equality deduction. (Contributed by Mario Carneiro, 2-Jan-2017.)
⊢ (𝜑 → 𝐴 = 𝐵)    &   ⊢ 𝐵 ⊆ 𝐶    ⇒   ⊢ (𝜑 → 𝐴 ⊆ 𝐶)
Theorem	eqsstrrdi 3985	A chained subclass and equality deduction. (Contributed by Mario Carneiro, 2-Jan-2017.)
⊢ (𝜑 → 𝐵 = 𝐴)    &   ⊢ 𝐵 ⊆ 𝐶    ⇒   ⊢ (𝜑 → 𝐴 ⊆ 𝐶)
Theorem	eqsstri 3986	Substitution of equality into a subclass relationship. (Contributed by NM, 16-Jul-1995.)
⊢ 𝐴 = 𝐵    &   ⊢ 𝐵 ⊆ 𝐶    ⇒   ⊢ 𝐴 ⊆ 𝐶
Theorem	eqsstrri 3987	Substitution of equality into a subclass relationship. (Contributed by NM, 19-Oct-1999.)
⊢ 𝐵 = 𝐴    &   ⊢ 𝐵 ⊆ 𝐶    ⇒   ⊢ 𝐴 ⊆ 𝐶
Theorem	sseqtri 3988	Substitution of equality into a subclass relationship. (Contributed by NM, 28-Jul-1995.)
⊢ 𝐴 ⊆ 𝐵    &   ⊢ 𝐵 = 𝐶    ⇒   ⊢ 𝐴 ⊆ 𝐶
Theorem	sseqtrri 3989	Substitution of equality into a subclass relationship. (Contributed by NM, 4-Apr-1995.)
⊢ 𝐴 ⊆ 𝐵    &   ⊢ 𝐶 = 𝐵    ⇒   ⊢ 𝐴 ⊆ 𝐶
Theorem	3sstr3i 3990	Substitution of equality in both sides of a subclass relationship. (Contributed by NM, 13-Jan-1996.) (Proof shortened by Eric Schmidt, 26-Jan-2007.)
⊢ 𝐴 ⊆ 𝐵    &   ⊢ 𝐴 = 𝐶    &   ⊢ 𝐵 = 𝐷    ⇒   ⊢ 𝐶 ⊆ 𝐷
Theorem	3sstr4i 3991	Substitution of equality in both sides of a subclass relationship. (Contributed by NM, 13-Jan-1996.) (Proof shortened by Eric Schmidt, 26-Jan-2007.)
⊢ 𝐴 ⊆ 𝐵    &   ⊢ 𝐶 = 𝐴    &   ⊢ 𝐷 = 𝐵    ⇒   ⊢ 𝐶 ⊆ 𝐷
Theorem	3sstr3g 3992	Substitution of equality into both sides of a subclass relationship. (Contributed by NM, 1-Oct-2000.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    &   ⊢ 𝐴 = 𝐶    &   ⊢ 𝐵 = 𝐷    ⇒   ⊢ (𝜑 → 𝐶 ⊆ 𝐷)
Theorem	3sstr4g 3993	Substitution of equality into both sides of a subclass relationship. (Contributed by NM, 16-Aug-1994.) (Proof shortened by Eric Schmidt, 26-Jan-2007.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    &   ⊢ 𝐶 = 𝐴    &   ⊢ 𝐷 = 𝐵    ⇒   ⊢ (𝜑 → 𝐶 ⊆ 𝐷)
Theorem	3sstr3d 3994	Substitution of equality into both sides of a subclass relationship. (Contributed by NM, 1-Oct-2000.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    &   ⊢ (𝜑 → 𝐴 = 𝐶)    &   ⊢ (𝜑 → 𝐵 = 𝐷)    ⇒   ⊢ (𝜑 → 𝐶 ⊆ 𝐷)
Theorem	3sstr4d 3995	Substitution of equality into both sides of a subclass relationship. (Contributed by NM, 30-Nov-1995.) (Proof shortened by Eric Schmidt, 26-Jan-2007.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    &   ⊢ (𝜑 → 𝐶 = 𝐴)    &   ⊢ (𝜑 → 𝐷 = 𝐵)    ⇒   ⊢ (𝜑 → 𝐶 ⊆ 𝐷)
Theorem	eqimssd 3996	Equality implies inclusion, deduction version. (Contributed by SN, 6-Nov-2024.)
⊢ (𝜑 → 𝐴 = 𝐵)    ⇒   ⊢ (𝜑 → 𝐴 ⊆ 𝐵)
Theorem	eqimsscd 3997	Equality implies inclusion, deduction version. (Contributed by SN, 15-Feb-2025.)
⊢ (𝜑 → 𝐴 = 𝐵)    ⇒   ⊢ (𝜑 → 𝐵 ⊆ 𝐴)
Theorem	eqimss 3998	Equality implies inclusion. (Contributed by NM, 21-Jun-1993.) (Proof shortened by Andrew Salmon, 21-Jun-2011.)
⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
Theorem	eqimss2 3999	Equality implies inclusion. (Contributed by NM, 23-Nov-2003.)
⊢ (𝐵 = 𝐴 → 𝐴 ⊆ 𝐵)
Theorem	eqimssi 4000	Infer subclass relationship from equality. (Contributed by NM, 6-Jan-2007.)
⊢ 𝐴 = 𝐵    ⇒   ⊢ 𝐴 ⊆ 𝐵