P. 45 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 4401-4500   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	abf 4401	A class abstraction determined by a false formula is empty. (Contributed by NM, 20-Jan-2012.) Avoid ax-8 2106, ax-10 2135, ax-11 2152, ax-12 2169. (Revised by Gino Giotto, 30-Jun-2024.)
⊢ ¬ 𝜑    ⇒   ⊢ {𝑥 ∣ 𝜑} = ∅
Theorem	abfOLD 4402	Obsolete version of abf 4401 as of 28-Jun-2024. (Contributed by NM, 20-Jan-2012.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ ¬ 𝜑    ⇒   ⊢ {𝑥 ∣ 𝜑} = ∅
Theorem	eq0rdv 4403*	Deduction for equality to the empty set. (Contributed by NM, 11-Jul-2014.) Avoid ax-8 2106, df-clel 2808. (Revised by Gino Giotto, 6-Sep-2024.)
⊢ (𝜑 → ¬ 𝑥 ∈ 𝐴)    ⇒   ⊢ (𝜑 → 𝐴 = ∅)
Theorem	eq0rdvALT 4404*	Alternate proof of eq0rdv 4403. Shorter, but requiring df-clel 2808, ax-8 2106. (Contributed by NM, 11-Jul-2014.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ (𝜑 → ¬ 𝑥 ∈ 𝐴)    ⇒   ⊢ (𝜑 → 𝐴 = ∅)
Theorem	csbprc 4405	The proper substitution of a proper class for a set into a class results in the empty set. (Contributed by NM, 17-Aug-2018.) (Proof shortened by JJ, 27-Aug-2021.)
⊢ (¬ 𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝐵 = ∅)
Theorem	csb0 4406	The proper substitution of a class into the empty set is the empty set. (Contributed by NM, 18-Aug-2018.)
⊢ ⦋𝐴 / 𝑥⦌∅ = ∅
Theorem	sbcel12 4407	Distribute proper substitution through a membership relation. (Contributed by NM, 10-Nov-2005.) (Revised by NM, 18-Aug-2018.)
⊢ ([𝐴 / 𝑥]𝐵 ∈ 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶)
Theorem	sbceqg 4408	Distribute proper substitution through an equality relation. (Contributed by NM, 10-Nov-2005.) (Proof shortened by Andrew Salmon, 29-Jun-2011.)
⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐶))
Theorem	sbceqi 4409	Distribution of class substitution over equality, in inference form. (Contributed by Giovanni Mascellani, 27-May-2019.)
⊢ 𝐴 ∈ V    &   ⊢ ⦋𝐴 / 𝑥⦌𝐵 = 𝐷    &   ⊢ ⦋𝐴 / 𝑥⦌𝐶 = 𝐸    ⇒   ⊢ ([𝐴 / 𝑥]𝐵 = 𝐶 ↔ 𝐷 = 𝐸)
Theorem	sbcnel12g 4410	Distribute proper substitution through negated membership. (Contributed by Andrew Salmon, 18-Jun-2011.)
⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 ∉ 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 ∉ ⦋𝐴 / 𝑥⦌𝐶))
Theorem	sbcne12 4411	Distribute proper substitution through an inequality. (Contributed by Andrew Salmon, 18-Jun-2011.) (Revised by NM, 18-Aug-2018.)
⊢ ([𝐴 / 𝑥]𝐵 ≠ 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 ≠ ⦋𝐴 / 𝑥⦌𝐶)
Theorem	sbcel1g 4412*	Move proper substitution in and out of a membership relation. Note that the scope of [𝐴 / 𝑥] is the wff 𝐵 ∈ 𝐶, whereas the scope of ⦋𝐴 / 𝑥⦌ is the class 𝐵. (Contributed by NM, 10-Nov-2005.)
⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 ∈ 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 ∈ 𝐶))
Theorem	sbceq1g 4413*	Move proper substitution to first argument of an equality. (Contributed by NM, 30-Nov-2005.)
⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 = 𝐶))
Theorem	sbcel2 4414*	Move proper substitution in and out of a membership relation. (Contributed by NM, 14-Nov-2005.) (Revised by NM, 18-Aug-2018.)
⊢ ([𝐴 / 𝑥]𝐵 ∈ 𝐶 ↔ 𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶)
Theorem	sbceq2g 4415*	Move proper substitution to second argument of an equality. (Contributed by NM, 30-Nov-2005.)
⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶 ↔ 𝐵 = ⦋𝐴 / 𝑥⦌𝐶))
Theorem	csbcom 4416*	Commutative law for double substitution into a class. (Contributed by NM, 14-Nov-2005.) (Revised by NM, 18-Aug-2018.)
⊢ ⦋𝐴 / 𝑥⦌⦋𝐵 / 𝑦⦌𝐶 = ⦋𝐵 / 𝑦⦌⦋𝐴 / 𝑥⦌𝐶
Theorem	sbcnestgfw 4417*	Nest the composition of two substitutions. Version of sbcnestgf 4422 with a disjoint variable condition, which does not require ax-13 2369. (Contributed by Mario Carneiro, 11-Nov-2016.) Avoid ax-13 2369. (Revised by Gino Giotto, 26-Jan-2024.)
⊢ ((𝐴 ∈ 𝑉 ∧ ∀𝑦Ⅎ𝑥𝜑) → ([𝐴 / 𝑥][𝐵 / 𝑦]𝜑 ↔ [⦋𝐴 / 𝑥⦌𝐵 / 𝑦]𝜑))
Theorem	csbnestgfw 4418*	Nest the composition of two substitutions. Version of csbnestgf 4423 with a disjoint variable condition, which does not require ax-13 2369. (Contributed by NM, 23-Nov-2005.) Avoid ax-13 2369. (Revised by Gino Giotto, 26-Jan-2024.)
⊢ ((𝐴 ∈ 𝑉 ∧ ∀𝑦Ⅎ𝑥𝐶) → ⦋𝐴 / 𝑥⦌⦋𝐵 / 𝑦⦌𝐶 = ⦋⦋𝐴 / 𝑥⦌𝐵 / 𝑦⦌𝐶)
Theorem	sbcnestgw 4419*	Nest the composition of two substitutions. Version of sbcnestg 4424 with a disjoint variable condition, which does not require ax-13 2369. (Contributed by NM, 27-Nov-2005.) Avoid ax-13 2369. (Revised by Gino Giotto, 26-Jan-2024.)
⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥][𝐵 / 𝑦]𝜑 ↔ [⦋𝐴 / 𝑥⦌𝐵 / 𝑦]𝜑))
Theorem	csbnestgw 4420*	Nest the composition of two substitutions. Version of csbnestg 4425 with a disjoint variable condition, which does not require ax-13 2369. (Contributed by NM, 23-Nov-2005.) Avoid ax-13 2369. (Revised by Gino Giotto, 26-Jan-2024.)
⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌⦋𝐵 / 𝑦⦌𝐶 = ⦋⦋𝐴 / 𝑥⦌𝐵 / 𝑦⦌𝐶)
Theorem	sbcco3gw 4421*	Composition of two substitutions. Version of sbcco3g 4426 with a disjoint variable condition, which does not require ax-13 2369. (Contributed by NM, 27-Nov-2005.) Avoid ax-13 2369. (Revised by Gino Giotto, 26-Jan-2024.)
⊢ (𝑥 = 𝐴 → 𝐵 = 𝐶)    ⇒   ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥][𝐵 / 𝑦]𝜑 ↔ [𝐶 / 𝑦]𝜑))
Theorem	sbcnestgf 4422	Nest the composition of two substitutions. Usage of this theorem is discouraged because it depends on ax-13 2369. Use the weaker sbcnestgfw 4417 when possible. (Contributed by Mario Carneiro, 11-Nov-2016.) (New usage is discouraged.)
⊢ ((𝐴 ∈ 𝑉 ∧ ∀𝑦Ⅎ𝑥𝜑) → ([𝐴 / 𝑥][𝐵 / 𝑦]𝜑 ↔ [⦋𝐴 / 𝑥⦌𝐵 / 𝑦]𝜑))
Theorem	csbnestgf 4423	Nest the composition of two substitutions. Usage of this theorem is discouraged because it depends on ax-13 2369. Use the weaker csbnestgfw 4418 when possible. (Contributed by NM, 23-Nov-2005.) (Proof shortened by Mario Carneiro, 10-Nov-2016.) (New usage is discouraged.)
⊢ ((𝐴 ∈ 𝑉 ∧ ∀𝑦Ⅎ𝑥𝐶) → ⦋𝐴 / 𝑥⦌⦋𝐵 / 𝑦⦌𝐶 = ⦋⦋𝐴 / 𝑥⦌𝐵 / 𝑦⦌𝐶)
Theorem	sbcnestg 4424*	Nest the composition of two substitutions. Usage of this theorem is discouraged because it depends on ax-13 2369. Use the weaker sbcnestgw 4419 when possible. (Contributed by NM, 27-Nov-2005.) (Proof shortened by Mario Carneiro, 11-Nov-2016.) (New usage is discouraged.)
⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥][𝐵 / 𝑦]𝜑 ↔ [⦋𝐴 / 𝑥⦌𝐵 / 𝑦]𝜑))
Theorem	csbnestg 4425*	Nest the composition of two substitutions. Usage of this theorem is discouraged because it depends on ax-13 2369. Use the weaker csbnestgw 4420 when possible. (Contributed by NM, 23-Nov-2005.) (Proof shortened by Mario Carneiro, 10-Nov-2016.) (New usage is discouraged.)
⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌⦋𝐵 / 𝑦⦌𝐶 = ⦋⦋𝐴 / 𝑥⦌𝐵 / 𝑦⦌𝐶)
Theorem	sbcco3g 4426*	Composition of two substitutions. Usage of this theorem is discouraged because it depends on ax-13 2369. Use the weaker sbcco3gw 4421 when possible. (Contributed by NM, 27-Nov-2005.) (Revised by Mario Carneiro, 11-Nov-2016.) (New usage is discouraged.)
⊢ (𝑥 = 𝐴 → 𝐵 = 𝐶)    ⇒   ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥][𝐵 / 𝑦]𝜑 ↔ [𝐶 / 𝑦]𝜑))
Theorem	csbco3g 4427*	Composition of two class substitutions. Usage of this theorem is discouraged because it depends on ax-13 2369. (Contributed by NM, 27-Nov-2005.) (Revised by Mario Carneiro, 11-Nov-2016.) (New usage is discouraged.)
⊢ (𝑥 = 𝐴 → 𝐵 = 𝐶)    ⇒   ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌⦋𝐵 / 𝑦⦌𝐷 = ⦋𝐶 / 𝑦⦌𝐷)
Theorem	csbnest1g 4428	Nest the composition of two substitutions. (Contributed by NM, 23-May-2006.) (Proof shortened by Mario Carneiro, 11-Nov-2016.)
⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌⦋𝐵 / 𝑥⦌𝐶 = ⦋⦋𝐴 / 𝑥⦌𝐵 / 𝑥⦌𝐶)
Theorem	csbidm 4429*	Idempotent law for class substitutions. (Contributed by NM, 1-Mar-2008.) (Revised by NM, 18-Aug-2018.)
⊢ ⦋𝐴 / 𝑥⦌⦋𝐴 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐵
Theorem	csbvarg 4430	The proper substitution of a class for setvar variable results in the class (if the class exists). (Contributed by NM, 10-Nov-2005.)
⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌𝑥 = 𝐴)
Theorem	csbvargi 4431	The proper substitution of a class for a setvar variable results in the class (if the class exists), in inference form of csbvarg 4430. (Contributed by Giovanni Mascellani, 30-May-2019.)
⊢ 𝐴 ∈ V    ⇒   ⊢ ⦋𝐴 / 𝑥⦌𝑥 = 𝐴
Theorem	sbccsb 4432*	Substitution into a wff expressed in terms of substitution into a class. (Contributed by NM, 15-Aug-2007.) (Revised by NM, 18-Aug-2018.)
⊢ ([𝐴 / 𝑥]𝜑 ↔ 𝑦 ∈ ⦋𝐴 / 𝑥⦌{𝑦 ∣ 𝜑})
Theorem	sbccsb2 4433	Substitution into a wff expressed in using substitution into a class. (Contributed by NM, 27-Nov-2005.) (Revised by NM, 18-Aug-2018.)
⊢ ([𝐴 / 𝑥]𝜑 ↔ 𝐴 ∈ ⦋𝐴 / 𝑥⦌{𝑥 ∣ 𝜑})
Theorem	rspcsbela 4434*	Special case related to rspsbc 3872. (Contributed by NM, 10-Dec-2005.) (Proof shortened by Eric Schmidt, 17-Jan-2007.)
⊢ ((𝐴 ∈ 𝐵 ∧ ∀𝑥 ∈ 𝐵 𝐶 ∈ 𝐷) → ⦋𝐴 / 𝑥⦌𝐶 ∈ 𝐷)
Theorem	sbnfc2 4435*	Two ways of expressing "𝑥 is (effectively) not free in 𝐴". (Contributed by Mario Carneiro, 14-Oct-2016.)
⊢ (Ⅎ𝑥𝐴 ↔ ∀𝑦∀𝑧⦋𝑦 / 𝑥⦌𝐴 = ⦋𝑧 / 𝑥⦌𝐴)
Theorem	csbab 4436*	Move substitution into a class abstraction. (Contributed by NM, 13-Dec-2005.) (Revised by NM, 19-Aug-2018.)
⊢ ⦋𝐴 / 𝑥⦌{𝑦 ∣ 𝜑} = {𝑦 ∣ [𝐴 / 𝑥]𝜑}
Theorem	csbun 4437	Distribution of class substitution over union of two classes. (Contributed by Drahflow, 23-Sep-2015.) (Revised by Mario Carneiro, 11-Dec-2016.) (Revised by NM, 13-Sep-2018.)
⊢ ⦋𝐴 / 𝑥⦌(𝐵 ∪ 𝐶) = (⦋𝐴 / 𝑥⦌𝐵 ∪ ⦋𝐴 / 𝑥⦌𝐶)
Theorem	csbin 4438	Distribute proper substitution into a class through an intersection relation. (Contributed by Alan Sare, 22-Jul-2012.) (Revised by NM, 18-Aug-2018.)
⊢ ⦋𝐴 / 𝑥⦌(𝐵 ∩ 𝐶) = (⦋𝐴 / 𝑥⦌𝐵 ∩ ⦋𝐴 / 𝑥⦌𝐶)
Theorem	csbie2df 4439*	Conversion of implicit substitution to explicit class substitution. This version of csbiedf 3923 avoids a disjointness condition on 𝑥, 𝐴 and 𝑥, 𝐷 by substituting twice. Deduction form of csbie2 3934. (Contributed by AV, 29-Mar-2024.)
⊢ Ⅎ𝑥𝜑    &   ⊢ (𝜑 → Ⅎ𝑥𝐶)    &   ⊢ (𝜑 → Ⅎ𝑥𝐷)    &   ⊢ (𝜑 → 𝐴 ∈ 𝑉)    &   ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → 𝐵 = 𝐶)    &   ⊢ ((𝜑 ∧ 𝑦 = 𝐴) → 𝐶 = 𝐷)    ⇒   ⊢ (𝜑 → ⦋𝐴 / 𝑥⦌𝐵 = 𝐷)
Theorem	2nreu 4440*	If there are two different sets fulfilling a wff (by implicit substitution), then there is no unique set fulfilling the wff. (Contributed by AV, 20-Jun-2023.)
⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))    &   ⊢ (𝑥 = 𝐵 → (𝜑 ↔ 𝜒))    ⇒   ⊢ ((𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ 𝐴 ≠ 𝐵) → ((𝜓 ∧ 𝜒) → ¬ ∃!𝑥 ∈ 𝑋 𝜑))
Theorem	un00 4441	Two classes are empty iff their union is empty. (Contributed by NM, 11-Aug-2004.)
⊢ ((𝐴 = ∅ ∧ 𝐵 = ∅) ↔ (𝐴 ∪ 𝐵) = ∅)
Theorem	vss 4442	Only the universal class has the universal class as a subclass. (Contributed by NM, 17-Sep-2003.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
⊢ (V ⊆ 𝐴 ↔ 𝐴 = V)
Theorem	0pss 4443	The null set is a proper subset of any nonempty set. (Contributed by NM, 27-Feb-1996.)
⊢ (∅ ⊊ 𝐴 ↔ 𝐴 ≠ ∅)
Theorem	npss0 4444	No set is a proper subset of the empty set. (Contributed by NM, 17-Jun-1998.) (Proof shortened by Andrew Salmon, 26-Jun-2011.) (Proof shortened by JJ, 14-Jul-2021.)
⊢ ¬ 𝐴 ⊊ ∅
Theorem	pssv 4445	Any non-universal class is a proper subclass of the universal class. (Contributed by NM, 17-May-1998.)
⊢ (𝐴 ⊊ V ↔ ¬ 𝐴 = V)
Theorem	disj 4446*	Two ways of saying that two classes are disjoint (have no members in common). (Contributed by NM, 17-Feb-2004.) Avoid ax-10 2135, ax-11 2152, ax-12 2169. (Revised by Gino Giotto, 28-Jun-2024.)
⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵)
Theorem	disjOLD 4447*	Obsolete version of disj 4446 as of 28-Jun-2024. (Contributed by NM, 17-Feb-2004.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵)
Theorem	disjr 4448*	Two ways of saying that two classes are disjoint. (Contributed by Jeff Madsen, 19-Jun-2011.)
⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥 ∈ 𝐵 ¬ 𝑥 ∈ 𝐴)
Theorem	disj1 4449*	Two ways of saying that two classes are disjoint (have no members in common). (Contributed by NM, 19-Aug-1993.)
⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵))
Theorem	reldisj 4450	Two ways of saying that two classes are disjoint, using the complement of 𝐵 relative to a universe 𝐶. (Contributed by NM, 15-Feb-2007.) (Proof shortened by Andrew Salmon, 26-Jun-2011.) Avoid ax-12 2169. (Revised by Gino Giotto, 28-Jun-2024.)
⊢ (𝐴 ⊆ 𝐶 → ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (𝐶 ∖ 𝐵)))
Theorem	reldisjOLD 4451	Obsolete version of reldisj 4450 as of 28-Jun-2024. (Contributed by NM, 15-Feb-2007.) (Proof shortened by Andrew Salmon, 26-Jun-2011.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ (𝐴 ⊆ 𝐶 → ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (𝐶 ∖ 𝐵)))
Theorem	disj3 4452	Two ways of saying that two classes are disjoint. (Contributed by NM, 19-May-1998.)
⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 = (𝐴 ∖ 𝐵))
Theorem	disjne 4453	Members of disjoint sets are not equal. (Contributed by NM, 28-Mar-2007.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ 𝐶 ∈ 𝐴 ∧ 𝐷 ∈ 𝐵) → 𝐶 ≠ 𝐷)
Theorem	disjeq0 4454	Two disjoint sets are equal iff both are empty. (Contributed by AV, 19-Jun-2022.)
⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 = 𝐵 ↔ (𝐴 = ∅ ∧ 𝐵 = ∅)))
Theorem	disjel 4455	A set can't belong to both members of disjoint classes. (Contributed by NM, 28-Feb-2015.)
⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ 𝐶 ∈ 𝐴) → ¬ 𝐶 ∈ 𝐵)
Theorem	disj2 4456	Two ways of saying that two classes are disjoint. (Contributed by NM, 17-May-1998.)
⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵))
Theorem	disj4 4457	Two ways of saying that two classes are disjoint. (Contributed by NM, 21-Mar-2004.)
⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ¬ (𝐴 ∖ 𝐵) ⊊ 𝐴)
Theorem	ssdisj 4458	Intersection with a subclass of a disjoint class. (Contributed by FL, 24-Jan-2007.) (Proof shortened by JJ, 14-Jul-2021.)
⊢ ((𝐴 ⊆ 𝐵 ∧ (𝐵 ∩ 𝐶) = ∅) → (𝐴 ∩ 𝐶) = ∅)
Theorem	disjpss 4459	A class is a proper subset of its union with a disjoint nonempty class. (Contributed by NM, 15-Sep-2004.)
⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ 𝐵 ≠ ∅) → 𝐴 ⊊ (𝐴 ∪ 𝐵))
Theorem	undisj1 4460	The union of disjoint classes is disjoint. (Contributed by NM, 26-Sep-2004.)
⊢ (((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐶) = ∅) ↔ ((𝐴 ∪ 𝐵) ∩ 𝐶) = ∅)
Theorem	undisj2 4461	The union of disjoint classes is disjoint. (Contributed by NM, 13-Sep-2004.)
⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ (𝐴 ∩ 𝐶) = ∅) ↔ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ∅)
Theorem	ssindif0 4462	Subclass expressed in terms of intersection with difference from the universal class. (Contributed by NM, 17-Sep-2003.)
⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ∩ (V ∖ 𝐵)) = ∅)
Theorem	inelcm 4463	The intersection of classes with a common member is nonempty. (Contributed by NM, 7-Apr-1994.)
⊢ ((𝐴 ∈ 𝐵 ∧ 𝐴 ∈ 𝐶) → (𝐵 ∩ 𝐶) ≠ ∅)
Theorem	minel 4464	A minimum element of a class has no elements in common with the class. (Contributed by NM, 22-Jun-1994.) (Proof shortened by JJ, 14-Jul-2021.)
⊢ ((𝐴 ∈ 𝐵 ∧ (𝐶 ∩ 𝐵) = ∅) → ¬ 𝐴 ∈ 𝐶)
Theorem	undif4 4465	Distribute union over difference. (Contributed by NM, 17-May-1998.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
⊢ ((𝐴 ∩ 𝐶) = ∅ → (𝐴 ∪ (𝐵 ∖ 𝐶)) = ((𝐴 ∪ 𝐵) ∖ 𝐶))
Theorem	disjssun 4466	Subset relation for disjoint classes. (Contributed by NM, 25-Oct-2005.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ 𝐴 ⊆ 𝐶))
Theorem	vdif0 4467	Universal class equality in terms of empty difference. (Contributed by NM, 17-Sep-2003.)
⊢ (𝐴 = V ↔ (V ∖ 𝐴) = ∅)
Theorem	difrab0eq 4468*	If the difference between the restricting class of a restricted class abstraction and the restricted class abstraction is empty, the restricting class is equal to this restricted class abstraction. (Contributed by Alexander van der Vekens, 31-Dec-2017.)
⊢ ((𝑉 ∖ {𝑥 ∈ 𝑉 ∣ 𝜑}) = ∅ ↔ 𝑉 = {𝑥 ∈ 𝑉 ∣ 𝜑})
Theorem	pssnel 4469*	A proper subclass has a member in one argument that's not in both. (Contributed by NM, 29-Feb-1996.)
⊢ (𝐴 ⊊ 𝐵 → ∃𝑥(𝑥 ∈ 𝐵 ∧ ¬ 𝑥 ∈ 𝐴))
Theorem	disjdif 4470	A class and its relative complement are disjoint. Theorem 38 of [Suppes] p. 29. (Contributed by NM, 24-Mar-1998.)
⊢ (𝐴 ∩ (𝐵 ∖ 𝐴)) = ∅
Theorem	disjdifr 4471	A class and its relative complement are disjoint. (Contributed by Thierry Arnoux, 29-Nov-2023.)
⊢ ((𝐵 ∖ 𝐴) ∩ 𝐴) = ∅
Theorem	difin0 4472	The difference of a class from its intersection is empty. Theorem 37 of [Suppes] p. 29. (Contributed by NM, 17-Aug-2004.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
⊢ ((𝐴 ∩ 𝐵) ∖ 𝐵) = ∅
Theorem	unvdif 4473	The union of a class and its complement is the universe. Theorem 5.1(5) of [Stoll] p. 17. (Contributed by NM, 17-Aug-2004.)
⊢ (𝐴 ∪ (V ∖ 𝐴)) = V
Theorem	undif1 4474	Absorption of difference by union. This decomposes a union into two disjoint classes (see disjdif 4470). Theorem 35 of [Suppes] p. 29. (Contributed by NM, 19-May-1998.)
⊢ ((𝐴 ∖ 𝐵) ∪ 𝐵) = (𝐴 ∪ 𝐵)
Theorem	undif2 4475	Absorption of difference by union. This decomposes a union into two disjoint classes (see disjdif 4470). Part of proof of Corollary 6K of [Enderton] p. 144. (Contributed by NM, 19-May-1998.)
⊢ (𝐴 ∪ (𝐵 ∖ 𝐴)) = (𝐴 ∪ 𝐵)
Theorem	undifabs 4476	Absorption of difference by union. (Contributed by NM, 18-Aug-2013.)
⊢ (𝐴 ∪ (𝐴 ∖ 𝐵)) = 𝐴
Theorem	inundif 4477	The intersection and class difference of a class with another class unite to give the original class. (Contributed by Paul Chapman, 5-Jun-2009.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
⊢ ((𝐴 ∩ 𝐵) ∪ (𝐴 ∖ 𝐵)) = 𝐴
Theorem	disjdif2 4478	The difference of a class and a class disjoint from it is the original class. (Contributed by BJ, 21-Apr-2019.)
⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ∖ 𝐵) = 𝐴)
Theorem	difun2 4479	Absorption of union by difference. Theorem 36 of [Suppes] p. 29. (Contributed by NM, 19-May-1998.)
⊢ ((𝐴 ∪ 𝐵) ∖ 𝐵) = (𝐴 ∖ 𝐵)
Theorem	undif 4480	Union of complementary parts into whole. (Contributed by NM, 22-Mar-1998.)
⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ∪ (𝐵 ∖ 𝐴)) = 𝐵)
Theorem	undifr 4481	Union of complementary parts into whole. (Contributed by Thierry Arnoux, 21-Nov-2023.) (Proof shortened by SN, 11-Mar-2025.)
⊢ (𝐴 ⊆ 𝐵 ↔ ((𝐵 ∖ 𝐴) ∪ 𝐴) = 𝐵)
Theorem	undifrOLD 4482	Obsolete version of undifr 4481 as of 11-Mar-2025. (Contributed by Thierry Arnoux, 21-Nov-2023.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ (𝐴 ⊆ 𝐵 ↔ ((𝐵 ∖ 𝐴) ∪ 𝐴) = 𝐵)
Theorem	undif5 4483	An equality involving class union and class difference. (Contributed by Thierry Arnoux, 26-Jun-2024.)
⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∪ 𝐵) ∖ 𝐵) = 𝐴)
Theorem	ssdifin0 4484	A subset of a difference does not intersect the subtrahend. (Contributed by Jeff Hankins, 1-Sep-2013.) (Proof shortened by Mario Carneiro, 24-Aug-2015.)
⊢ (𝐴 ⊆ (𝐵 ∖ 𝐶) → (𝐴 ∩ 𝐶) = ∅)
Theorem	ssdifeq0 4485	A class is a subclass of itself subtracted from another iff it is the empty set. (Contributed by Steve Rodriguez, 20-Nov-2015.)
⊢ (𝐴 ⊆ (𝐵 ∖ 𝐴) ↔ 𝐴 = ∅)
Theorem	ssundif 4486	A condition equivalent to inclusion in the union of two classes. (Contributed by NM, 26-Mar-2007.)
⊢ (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ (𝐴 ∖ 𝐵) ⊆ 𝐶)
Theorem	difcom 4487	Swap the arguments of a class difference. (Contributed by NM, 29-Mar-2007.)
⊢ ((𝐴 ∖ 𝐵) ⊆ 𝐶 ↔ (𝐴 ∖ 𝐶) ⊆ 𝐵)
Theorem	pssdifcom1 4488	Two ways to express overlapping subsets. (Contributed by Stefan O'Rear, 31-Oct-2014.)
⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → ((𝐶 ∖ 𝐴) ⊊ 𝐵 ↔ (𝐶 ∖ 𝐵) ⊊ 𝐴))
Theorem	pssdifcom2 4489	Two ways to express non-covering pairs of subsets. (Contributed by Stefan O'Rear, 31-Oct-2014.)
⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶) → (𝐵 ⊊ (𝐶 ∖ 𝐴) ↔ 𝐴 ⊊ (𝐶 ∖ 𝐵)))
Theorem	difdifdir 4490	Distributive law for class difference. Exercise 4.8 of [Stoll] p. 16. (Contributed by NM, 18-Aug-2004.)
⊢ ((𝐴 ∖ 𝐵) ∖ 𝐶) = ((𝐴 ∖ 𝐶) ∖ (𝐵 ∖ 𝐶))
Theorem	uneqdifeq 4491	Two ways to say that 𝐴 and 𝐵 partition 𝐶 (when 𝐴 and 𝐵 don't overlap and 𝐴 is a part of 𝐶). (Contributed by FL, 17-Nov-2008.) (Proof shortened by JJ, 14-Jul-2021.)
⊢ ((𝐴 ⊆ 𝐶 ∧ (𝐴 ∩ 𝐵) = ∅) → ((𝐴 ∪ 𝐵) = 𝐶 ↔ (𝐶 ∖ 𝐴) = 𝐵))
Theorem	raldifeq 4492*	Equality theorem for restricted universal quantifier. (Contributed by Thierry Arnoux, 6-Jul-2019.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    &   ⊢ (𝜑 → ∀𝑥 ∈ (𝐵 ∖ 𝐴)𝜓)    ⇒   ⊢ (𝜑 → (∀𝑥 ∈ 𝐴 𝜓 ↔ ∀𝑥 ∈ 𝐵 𝜓))
Theorem	r19.2z 4493*	Theorem 19.2 of [Margaris] p. 89 with restricted quantifiers (compare 19.2 1978). The restricted version is valid only when the domain of quantification is not empty. (Contributed by NM, 15-Nov-2003.)
⊢ ((𝐴 ≠ ∅ ∧ ∀𝑥 ∈ 𝐴 𝜑) → ∃𝑥 ∈ 𝐴 𝜑)
Theorem	r19.2zb 4494*	A response to the notion that the condition 𝐴 ≠ ∅ can be removed in r19.2z 4493. Interestingly enough, 𝜑 does not figure in the left-hand side. (Contributed by Jeff Hankins, 24-Aug-2009.)
⊢ (𝐴 ≠ ∅ ↔ (∀𝑥 ∈ 𝐴 𝜑 → ∃𝑥 ∈ 𝐴 𝜑))
Theorem	r19.3rz 4495*	Restricted quantification of wff not containing quantified variable. (Contributed by FL, 3-Jan-2008.)
⊢ Ⅎ𝑥𝜑    ⇒   ⊢ (𝐴 ≠ ∅ → (𝜑 ↔ ∀𝑥 ∈ 𝐴 𝜑))
Theorem	r19.28z 4496*	Restricted quantifier version of Theorem 19.28 of [Margaris] p. 90. It is valid only when the domain of quantification is not empty. (Contributed by NM, 26-Oct-2010.)
⊢ Ⅎ𝑥𝜑    ⇒   ⊢ (𝐴 ≠ ∅ → (∀𝑥 ∈ 𝐴 (𝜑 ∧ 𝜓) ↔ (𝜑 ∧ ∀𝑥 ∈ 𝐴 𝜓)))
Theorem	r19.3rzv 4497*	Restricted quantification of wff not containing quantified variable. (Contributed by NM, 10-Mar-1997.)
⊢ (𝐴 ≠ ∅ → (𝜑 ↔ ∀𝑥 ∈ 𝐴 𝜑))
Theorem	r19.9rzv 4498*	Restricted quantification of wff not containing quantified variable. (Contributed by NM, 27-May-1998.)
⊢ (𝐴 ≠ ∅ → (𝜑 ↔ ∃𝑥 ∈ 𝐴 𝜑))
Theorem	r19.28zv 4499*	Restricted quantifier version of Theorem 19.28 of [Margaris] p. 90. It is valid only when the domain of quantification is not empty. (Contributed by NM, 19-Aug-2004.)
⊢ (𝐴 ≠ ∅ → (∀𝑥 ∈ 𝐴 (𝜑 ∧ 𝜓) ↔ (𝜑 ∧ ∀𝑥 ∈ 𝐴 𝜓)))
Theorem	r19.37zv 4500*	Restricted quantifier version of Theorem 19.37 of [Margaris] p. 90. It is valid only when the domain of quantification is not empty. (Contributed by Paul Chapman, 8-Oct-2007.)
⊢ (𝐴 ≠ ∅ → (∃𝑥 ∈ 𝐴 (𝜑 → 𝜓) ↔ (𝜑 → ∃𝑥 ∈ 𝐴 𝜓)))