Home Metamath Proof ExplorerTheorem List (p. 41 of 424) < Previous  Next > Bad symbols? Try the GIF version. Mirrors  >  Metamath Home Page  >  MPE Home Page  >  Theorem List Contents  >  Recent Proofs       This page: Page List

 Color key: Metamath Proof Explorer (1-27759) Hilbert Space Explorer (27760-29284) Users' Mathboxes (29285-42322)

Theorem List for Metamath Proof Explorer - 4001-4100   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremcsbvarg 4001 The proper substitution of a class for setvar variable results in the class (if the class exists). (Contributed by NM, 10-Nov-2005.)
(𝐴𝑉𝐴 / 𝑥𝑥 = 𝐴)

Theoremsbccsb 4002* 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]𝜑𝑦𝐴 / 𝑥{𝑦𝜑})

Theoremsbccsb2 4003 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]𝜑𝐴𝐴 / 𝑥{𝑥𝜑})

Theoremrspcsbela 4004* Special case related to rspsbc 3516. (Contributed by NM, 10-Dec-2005.) (Proof shortened by Eric Schmidt, 17-Jan-2007.)
((𝐴𝐵 ∧ ∀𝑥𝐵 𝐶𝐷) → 𝐴 / 𝑥𝐶𝐷)

Theoremsbnfc2 4005* Two ways of expressing "𝑥 is (effectively) not free in 𝐴." (Contributed by Mario Carneiro, 14-Oct-2016.)
(𝑥𝐴 ↔ ∀𝑦𝑧𝑦 / 𝑥𝐴 = 𝑧 / 𝑥𝐴)

Theoremcsbab 4006* Move substitution into a class abstraction. (Contributed by NM, 13-Dec-2005.) (Revised by NM, 19-Aug-2018.)
𝐴 / 𝑥{𝑦𝜑} = {𝑦[𝐴 / 𝑥]𝜑}

Theoremcsbun 4007 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(𝐵𝐶) = (𝐴 / 𝑥𝐵𝐴 / 𝑥𝐶)

Theoremcsbin 4008 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(𝐵𝐶) = (𝐴 / 𝑥𝐵𝐴 / 𝑥𝐶)

Theoremun00 4009 Two classes are empty iff their union is empty. (Contributed by NM, 11-Aug-2004.)
((𝐴 = ∅ ∧ 𝐵 = ∅) ↔ (𝐴𝐵) = ∅)

Theoremvss 4010 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 ⊆ 𝐴𝐴 = V)

Theorem0pss 4011 The null set is a proper subset of any nonempty set. (Contributed by NM, 27-Feb-1996.)
(∅ ⊊ 𝐴𝐴 ≠ ∅)

Theoremnpss0 4012 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 ⊊ ∅

Theoremnpss0OLD 4013 Obsolete proof of npss0 4012 as of 14-Jul-2021. (Contributed by NM, 17-Jun-1998.) (Proof shortened by Andrew Salmon, 26-Jun-2011.) (New usage is discouraged.) (Proof modification is discouraged.)
¬ 𝐴 ⊊ ∅

Theorempssv 4014 Any non-universal class is a proper subclass of the universal class. (Contributed by NM, 17-May-1998.)
(𝐴 ⊊ V ↔ ¬ 𝐴 = V)

Theoremdisj 4015* Two ways of saying that two classes are disjoint (have no members in common). (Contributed by NM, 17-Feb-2004.)
((𝐴𝐵) = ∅ ↔ ∀𝑥𝐴 ¬ 𝑥𝐵)

Theoremdisjr 4016* Two ways of saying that two classes are disjoint. (Contributed by Jeff Madsen, 19-Jun-2011.)
((𝐴𝐵) = ∅ ↔ ∀𝑥𝐵 ¬ 𝑥𝐴)

Theoremdisj1 4017* Two ways of saying that two classes are disjoint (have no members in common). (Contributed by NM, 19-Aug-1993.)
((𝐴𝐵) = ∅ ↔ ∀𝑥(𝑥𝐴 → ¬ 𝑥𝐵))

Theoremreldisj 4018 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.)
(𝐴𝐶 → ((𝐴𝐵) = ∅ ↔ 𝐴 ⊆ (𝐶𝐵)))

Theoremdisj3 4019 Two ways of saying that two classes are disjoint. (Contributed by NM, 19-May-1998.)
((𝐴𝐵) = ∅ ↔ 𝐴 = (𝐴𝐵))

Theoremdisjne 4020 Members of disjoint sets are not equal. (Contributed by NM, 28-Mar-2007.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
(((𝐴𝐵) = ∅ ∧ 𝐶𝐴𝐷𝐵) → 𝐶𝐷)

Theoremdisjel 4021 A set can't belong to both members of disjoint classes. (Contributed by NM, 28-Feb-2015.)
(((𝐴𝐵) = ∅ ∧ 𝐶𝐴) → ¬ 𝐶𝐵)

Theoremdisj2 4022 Two ways of saying that two classes are disjoint. (Contributed by NM, 17-May-1998.)
((𝐴𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵))

Theoremdisj4 4023 Two ways of saying that two classes are disjoint. (Contributed by NM, 21-Mar-2004.)
((𝐴𝐵) = ∅ ↔ ¬ (𝐴𝐵) ⊊ 𝐴)

Theoremssdisj 4024 Intersection with a subclass of a disjoint class. (Contributed by FL, 24-Jan-2007.) (Proof shortened by JJ, 14-Jul-2021.)
((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) = ∅)

TheoremssdisjOLD 4025 Obsolete proof of ssdisj 4024 as of 14-Jul-2021. (Contributed by FL, 24-Jan-2007.) (New usage is discouraged.) (Proof modification is discouraged.)
((𝐴𝐵 ∧ (𝐵𝐶) = ∅) → (𝐴𝐶) = ∅)

Theoremdisjpss 4026 A class is a proper subset of its union with a disjoint nonempty class. (Contributed by NM, 15-Sep-2004.)
(((𝐴𝐵) = ∅ ∧ 𝐵 ≠ ∅) → 𝐴 ⊊ (𝐴𝐵))

Theoremundisj1 4027 The union of disjoint classes is disjoint. (Contributed by NM, 26-Sep-2004.)
(((𝐴𝐶) = ∅ ∧ (𝐵𝐶) = ∅) ↔ ((𝐴𝐵) ∩ 𝐶) = ∅)

Theoremundisj2 4028 The union of disjoint classes is disjoint. (Contributed by NM, 13-Sep-2004.)
(((𝐴𝐵) = ∅ ∧ (𝐴𝐶) = ∅) ↔ (𝐴 ∩ (𝐵𝐶)) = ∅)

Theoremssindif0 4029 Subclass expressed in terms of intersection with difference from the universal class. (Contributed by NM, 17-Sep-2003.)
(𝐴𝐵 ↔ (𝐴 ∩ (V ∖ 𝐵)) = ∅)

Theoreminelcm 4030 The intersection of classes with a common member is nonempty. (Contributed by NM, 7-Apr-1994.)
((𝐴𝐵𝐴𝐶) → (𝐵𝐶) ≠ ∅)

Theoremminel 4031 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.)
((𝐴𝐵 ∧ (𝐶𝐵) = ∅) → ¬ 𝐴𝐶)

TheoremminelOLD 4032 Obsolete proof of minel 4031 as of 14-Jul-2021. (Contributed by NM, 22-Jun-1994.) (New usage is discouraged.) (Proof modification is discouraged.)
((𝐴𝐵 ∧ (𝐶𝐵) = ∅) → ¬ 𝐴𝐶)

Theoremundif4 4033 Distribute union over difference. (Contributed by NM, 17-May-1998.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
((𝐴𝐶) = ∅ → (𝐴 ∪ (𝐵𝐶)) = ((𝐴𝐵) ∖ 𝐶))

Theoremdisjssun 4034 Subset relation for disjoint classes. (Contributed by NM, 25-Oct-2005.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
((𝐴𝐵) = ∅ → (𝐴 ⊆ (𝐵𝐶) ↔ 𝐴𝐶))

Theoremvdif0 4035 Universal class equality in terms of empty difference. (Contributed by NM, 17-Sep-2003.)
(𝐴 = V ↔ (V ∖ 𝐴) = ∅)

Theoremdifrab0eq 4036* 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 ∖ {𝑥𝑉𝜑}) = ∅ ↔ 𝑉 = {𝑥𝑉𝜑})

Theorempssnel 4037* A proper subclass has a member in one argument that's not in both. (Contributed by NM, 29-Feb-1996.)
(𝐴𝐵 → ∃𝑥(𝑥𝐵 ∧ ¬ 𝑥𝐴))

Theoremdisjdif 4038 A class and its relative complement are disjoint. Theorem 38 of [Suppes] p. 29. (Contributed by NM, 24-Mar-1998.)
(𝐴 ∩ (𝐵𝐴)) = ∅

Theoremdifin0 4039 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.)
((𝐴𝐵) ∖ 𝐵) = ∅

Theoremunvdif 4040 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

Theoremundif1 4041 Absorption of difference by union. This decomposes a union into two disjoint classes (see disjdif 4038). Theorem 35 of [Suppes] p. 29. (Contributed by NM, 19-May-1998.)
((𝐴𝐵) ∪ 𝐵) = (𝐴𝐵)

Theoremundif2 4042 Absorption of difference by union. This decomposes a union into two disjoint classes (see disjdif 4038). Part of proof of Corollary 6K of [Enderton] p. 144. (Contributed by NM, 19-May-1998.)
(𝐴 ∪ (𝐵𝐴)) = (𝐴𝐵)

Theoremundifabs 4043 Absorption of difference by union. (Contributed by NM, 18-Aug-2013.)
(𝐴 ∪ (𝐴𝐵)) = 𝐴

Theoreminundif 4044 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.)
((𝐴𝐵) ∪ (𝐴𝐵)) = 𝐴

Theoremdisjdif2 4045 The difference of a class and a class disjoint from it is the original class. (Contributed by BJ, 21-Apr-2019.)
((𝐴𝐵) = ∅ → (𝐴𝐵) = 𝐴)

Theoremdifun2 4046 Absorption of union by difference. Theorem 36 of [Suppes] p. 29. (Contributed by NM, 19-May-1998.)
((𝐴𝐵) ∖ 𝐵) = (𝐴𝐵)

Theoremundif 4047 Union of complementary parts into whole. (Contributed by NM, 22-Mar-1998.)
(𝐴𝐵 ↔ (𝐴 ∪ (𝐵𝐴)) = 𝐵)

Theoremssdifin0 4048 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 ⊆ (𝐵𝐶) → (𝐴𝐶) = ∅)

Theoremssdifeq0 4049 A class is a subclass of itself subtracted from another iff it is the empty set. (Contributed by Steve Rodriguez, 20-Nov-2015.)
(𝐴 ⊆ (𝐵𝐴) ↔ 𝐴 = ∅)

Theoremssundif 4050 A condition equivalent to inclusion in the union of two classes. (Contributed by NM, 26-Mar-2007.)
(𝐴 ⊆ (𝐵𝐶) ↔ (𝐴𝐵) ⊆ 𝐶)

Theoremdifcom 4051 Swap the arguments of a class difference. (Contributed by NM, 29-Mar-2007.)
((𝐴𝐵) ⊆ 𝐶 ↔ (𝐴𝐶) ⊆ 𝐵)

Theorempssdifcom1 4052 Two ways to express overlapping subsets. (Contributed by Stefan O'Rear, 31-Oct-2014.)
((𝐴𝐶𝐵𝐶) → ((𝐶𝐴) ⊊ 𝐵 ↔ (𝐶𝐵) ⊊ 𝐴))

Theorempssdifcom2 4053 Two ways to express non-covering pairs of subsets. (Contributed by Stefan O'Rear, 31-Oct-2014.)
((𝐴𝐶𝐵𝐶) → (𝐵 ⊊ (𝐶𝐴) ↔ 𝐴 ⊊ (𝐶𝐵)))

Theoremdifdifdir 4054 Distributive law for class difference. Exercise 4.8 of [Stoll] p. 16. (Contributed by NM, 18-Aug-2004.)
((𝐴𝐵) ∖ 𝐶) = ((𝐴𝐶) ∖ (𝐵𝐶))

Theoremuneqdifeq 4055 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.)
((𝐴𝐶 ∧ (𝐴𝐵) = ∅) → ((𝐴𝐵) = 𝐶 ↔ (𝐶𝐴) = 𝐵))

TheoremuneqdifeqOLD 4056 Obsolete proof of uneqdifeq 4055 as of 14-Jul-2021. (Contributed by FL, 17-Nov-2008.) (New usage is discouraged.) (Proof modification is discouraged.)
((𝐴𝐶 ∧ (𝐴𝐵) = ∅) → ((𝐴𝐵) = 𝐶 ↔ (𝐶𝐴) = 𝐵))

Theoremraldifeq 4057* Equality theorem for restricted universal quantifier. (Contributed by Thierry Arnoux, 6-Jul-2019.)
(𝜑𝐴𝐵)    &   (𝜑 → ∀𝑥 ∈ (𝐵𝐴)𝜓)       (𝜑 → (∀𝑥𝐴 𝜓 ↔ ∀𝑥𝐵 𝜓))

Theoremr19.2z 4058* Theorem 19.2 of [Margaris] p. 89 with restricted quantifiers (compare 19.2 1891). The restricted version is valid only when the domain of quantification is not empty. (Contributed by NM, 15-Nov-2003.)
((𝐴 ≠ ∅ ∧ ∀𝑥𝐴 𝜑) → ∃𝑥𝐴 𝜑)

Theoremr19.2zb 4059* A response to the notion that the condition 𝐴 ≠ ∅ can be removed in r19.2z 4058. Interestingly enough, 𝜑 does not figure in the left-hand side. (Contributed by Jeff Hankins, 24-Aug-2009.)
(𝐴 ≠ ∅ ↔ (∀𝑥𝐴 𝜑 → ∃𝑥𝐴 𝜑))

Theoremr19.3rz 4060* Restricted quantification of wff not containing quantified variable. (Contributed by FL, 3-Jan-2008.)
𝑥𝜑       (𝐴 ≠ ∅ → (𝜑 ↔ ∀𝑥𝐴 𝜑))

Theoremr19.28z 4061* 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.)
𝑥𝜑       (𝐴 ≠ ∅ → (∀𝑥𝐴 (𝜑𝜓) ↔ (𝜑 ∧ ∀𝑥𝐴 𝜓)))

Theoremr19.3rzv 4062* Restricted quantification of wff not containing quantified variable. (Contributed by NM, 10-Mar-1997.)
(𝐴 ≠ ∅ → (𝜑 ↔ ∀𝑥𝐴 𝜑))

Theoremr19.9rzv 4063* Restricted quantification of wff not containing quantified variable. (Contributed by NM, 27-May-1998.)
(𝐴 ≠ ∅ → (𝜑 ↔ ∃𝑥𝐴 𝜑))

Theoremr19.28zv 4064* 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 ≠ ∅ → (∀𝑥𝐴 (𝜑𝜓) ↔ (𝜑 ∧ ∀𝑥𝐴 𝜓)))

Theoremr19.37zv 4065* 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 ≠ ∅ → (∃𝑥𝐴 (𝜑𝜓) ↔ (𝜑 → ∃𝑥𝐴 𝜓)))

Theoremr19.45zv 4066* Restricted version of Theorem 19.45 of [Margaris] p. 90. (Contributed by NM, 27-May-1998.)
(𝐴 ≠ ∅ → (∃𝑥𝐴 (𝜑𝜓) ↔ (𝜑 ∨ ∃𝑥𝐴 𝜓)))

Theoremr19.44zv 4067* Restricted version of Theorem 19.44 of [Margaris] p. 90. (Contributed by NM, 27-May-1998.)
(𝐴 ≠ ∅ → (∃𝑥𝐴 (𝜑𝜓) ↔ (∃𝑥𝐴 𝜑𝜓)))

Theoremr19.27z 4068* Restricted quantifier version of Theorem 19.27 of [Margaris] p. 90. It is valid only when the domain of quantification is not empty. (Contributed by NM, 26-Oct-2010.)
𝑥𝜓       (𝐴 ≠ ∅ → (∀𝑥𝐴 (𝜑𝜓) ↔ (∀𝑥𝐴 𝜑𝜓)))

Theoremr19.27zv 4069* Restricted quantifier version of Theorem 19.27 of [Margaris] p. 90. It is valid only when the domain of quantification is not empty. (Contributed by NM, 19-Aug-2004.)
(𝐴 ≠ ∅ → (∀𝑥𝐴 (𝜑𝜓) ↔ (∀𝑥𝐴 𝜑𝜓)))

Theoremr19.36zv 4070* Restricted quantifier version of Theorem 19.36 of [Margaris] p. 90. It is valid only when the domain of quantification is not empty. (Contributed by NM, 20-Sep-2003.)
(𝐴 ≠ ∅ → (∃𝑥𝐴 (𝜑𝜓) ↔ (∀𝑥𝐴 𝜑𝜓)))

Theoremrzal 4071* Vacuous quantification is always true. (Contributed by NM, 11-Mar-1997.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
(𝐴 = ∅ → ∀𝑥𝐴 𝜑)

Theoremrexn0 4072* Restricted existential quantification implies its restriction is nonempty. (Contributed by Szymon Jaroszewicz, 3-Apr-2007.)
(∃𝑥𝐴 𝜑𝐴 ≠ ∅)

Theoremralidm 4073* Idempotent law for restricted quantifier. (Contributed by NM, 28-Mar-1997.)
(∀𝑥𝐴𝑥𝐴 𝜑 ↔ ∀𝑥𝐴 𝜑)

Theoremral0 4074 Vacuous universal quantification is always true. (Contributed by NM, 20-Oct-2005.)
𝑥 ∈ ∅ 𝜑

TheoremrgenzOLD 4075* Obsolete as of 22-Jul-2021. (Contributed by NM, 8-Dec-2012.) (New usage is discouraged.) (Proof modification is discouraged.)
((𝐴 ≠ ∅ ∧ 𝑥𝐴) → 𝜑)       𝑥𝐴 𝜑

Theoremralf0 4076* The quantification of a falsehood is vacuous when true. (Contributed by NM, 26-Nov-2005.) (Proof shortened by JJ, 14-Jul-2021.)
¬ 𝜑       (∀𝑥𝐴 𝜑𝐴 = ∅)

Theoremralf0OLD 4077* Obsolete proof of ralf0 4076 as of 14-Jul-2021. (Contributed by NM, 26-Nov-2005.) (New usage is discouraged.) (Proof modification is discouraged.)
¬ 𝜑       (∀𝑥𝐴 𝜑𝐴 = ∅)

Theoremralnralall 4078* A contradiction concerning restricted generalization for a nonempty set implies anything. (Contributed by Alexander van der Vekens, 4-Sep-2018.)
(𝐴 ≠ ∅ → ((∀𝑥𝐴 𝜑 ∧ ∀𝑥𝐴 ¬ 𝜑) → 𝜓))

Theoremfalseral0 4079* A false statement can only be true for elements of an empty set. (Contributed by AV, 30-Oct-2020.)
((∀𝑥 ¬ 𝜑 ∧ ∀𝑥𝐴 𝜑) → 𝐴 = ∅)

Theoremraaan 4080* Rearrange restricted quantifiers. (Contributed by NM, 26-Oct-2010.)
𝑦𝜑    &   𝑥𝜓       (∀𝑥𝐴𝑦𝐴 (𝜑𝜓) ↔ (∀𝑥𝐴 𝜑 ∧ ∀𝑦𝐴 𝜓))

Theoremraaanv 4081* Rearrange restricted quantifiers. (Contributed by NM, 11-Mar-1997.)
(∀𝑥𝐴𝑦𝐴 (𝜑𝜓) ↔ (∀𝑥𝐴 𝜑 ∧ ∀𝑦𝐴 𝜓))

Theoremsbss 4082* Set substitution into the first argument of a subset relation. (Contributed by Rodolfo Medina, 7-Jul-2010.) (Proof shortened by Mario Carneiro, 14-Nov-2016.)
([𝑦 / 𝑥]𝑥𝐴𝑦𝐴)

Theoremsbcssg 4083 Distribute proper substitution through a subclass relation. (Contributed by Alan Sare, 22-Jul-2012.) (Proof shortened by Alexander van der Vekens, 23-Jul-2017.)
(𝐴𝑉 → ([𝐴 / 𝑥]𝐵𝐶𝐴 / 𝑥𝐵𝐴 / 𝑥𝐶))

2.1.15  "Weak deduction theorem" for set theory

In a Hilbert system of logic (which consists of a set of axioms, modus ponens, and the generalization rule), converting a deduction to a proof using the Deduction Theorem (taught in introductory logic books) involves an exponential increase of the number of steps as hypotheses are successively eliminated. Here is a trick that is not as general as the Deduction Theorem but requires only a linear increase in the number of steps.

The general problem: We want to convert a deduction P |- Q into a proof of the theorem |- P -> Q i.e., we want to eliminate the hypothesis P. Normally this is done using the Deduction (meta)Theorem, which looks at the microscopic steps of the deduction and usually doubles or triples the number of these microscopic steps for each hypothesis that is eliminated. We will look at a special case of this problem, without appealing to the Deduction Theorem.

We assume ZF with class notation. A and B are arbitrary (possibly proper) classes. P, Q, R, S and T are wffs.

We define the conditional operator, if(P, A, B), as follows: if(P, A, B) =def= { x | (x \in A & P) v (x \in B & -. P) } (where x does not occur in A, B, or P).

Lemma 1. A = if(P, A, B) -> (P <-> R), B = if(P, A, B) -> (S <-> R), S |- R Proof: Logic and Axiom of Extensionality.

Lemma 2. A = if(P, A, B) -> (Q <-> T), T |- P -> Q Proof: Logic and Axiom of Extensionality.

Here's a simple example that illustrates how it works. Suppose we have a deduction Ord A |- Tr A which means, "Assume A is an ordinal class. Then A is a transitive class." Note that A is a class variable that may be substituted with any class expression, so this is really a deduction scheme.

We want to convert this to a proof of the theorem (scheme) |- Ord A -> Tr A.

The catch is that we must be able to prove "Ord A" for at least one object A (and this is what makes it weaker than the ordinary Deduction Theorem). However, it is easy to prove |- Ord 0 (the empty set is ordinal). (For a typical textbook "theorem," i.e. deduction, there is usually at least one object satisfying each hypothesis, otherwise the theorem would not be very useful. We can always go back to the standard Deduction Theorem for those hypotheses where this is not the case.) Continuing with the example:

Equality axioms (and Extensionality) yield |- A = if(Ord A, A, 0) -> (Ord A <-> Ord if(Ord A, A, 0)) (1) |- 0 = if(Ord A, A, 0) -> (Ord 0 <-> Ord if(Ord A, A, 0)) (2) From (1), (2) and |- Ord 0, Lemma 1 yields |- Ord if(Ord A, A, 0) (3) From (3) and substituting if(Ord A, A, 0) for A in the original deduction, |- Tr if(Ord A, A, 0) (4) Equality axioms (and Extensionality) yield |- A = if(Ord A, A, 0) -> (Tr A <-> Tr if(Ord A, A, 0)) (5) From (4) and (5), Lemma 2 yields |- Ord A -> Tr A (Q.E.D.)

Syntaxcif 4084 Extend class notation to include the conditional operator. See df-if 4085 for a description. (In older databases this was denoted "ded".)
class if(𝜑, 𝐴, 𝐵)

Definitiondf-if 4085* Define the conditional operator. Read if(𝜑, 𝐴, 𝐵) as "if 𝜑 then 𝐴 else 𝐵." See iftrue 4090 and iffalse 4093 for its values. In mathematical literature, this operator is rarely defined formally but is implicit in informal definitions such as "let f(x)=0 if x=0 and 1/x otherwise." (In older versions of this database, this operator was denoted "ded" and called the "deduction class.")

An important use for us is in conjunction with the weak deduction theorem, which converts a hypothesis into an antecedent. In that role, 𝐴 is a class variable in the hypothesis and 𝐵 is a class (usually a constant) that makes the hypothesis true when it is substituted for 𝐴. See dedth 4137 for the main part of the weak deduction theorem, elimhyp 4144 to eliminate a hypothesis, and keephyp 4150 to keep a hypothesis. See the Deduction Theorem link on the Metamath Proof Explorer Home Page for a description of the weak deduction theorem. (Contributed by NM, 15-May-1999.)

if(𝜑, 𝐴, 𝐵) = {𝑥 ∣ ((𝑥𝐴𝜑) ∨ (𝑥𝐵 ∧ ¬ 𝜑))}

Theoremdfif2 4086* An alternate definition of the conditional operator df-if 4085 with one fewer connectives (but probably less intuitive to understand). (Contributed by NM, 30-Jan-2006.)
if(𝜑, 𝐴, 𝐵) = {𝑥 ∣ ((𝑥𝐵𝜑) → (𝑥𝐴𝜑))}

Theoremdfif6 4087* An alternate definition of the conditional operator df-if 4085 as a simple class abstraction. (Contributed by Mario Carneiro, 8-Sep-2013.)
if(𝜑, 𝐴, 𝐵) = ({𝑥𝐴𝜑} ∪ {𝑥𝐵 ∣ ¬ 𝜑})

Theoremifeq1 4088 Equality theorem for conditional operator. (Contributed by NM, 1-Sep-2004.) (Revised by Mario Carneiro, 8-Sep-2013.)
(𝐴 = 𝐵 → if(𝜑, 𝐴, 𝐶) = if(𝜑, 𝐵, 𝐶))

Theoremifeq2 4089 Equality theorem for conditional operator. (Contributed by NM, 1-Sep-2004.) (Revised by Mario Carneiro, 8-Sep-2013.)
(𝐴 = 𝐵 → if(𝜑, 𝐶, 𝐴) = if(𝜑, 𝐶, 𝐵))

Theoremiftrue 4090 Value of the conditional operator when its first argument is true. (Contributed by NM, 15-May-1999.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
(𝜑 → if(𝜑, 𝐴, 𝐵) = 𝐴)

Theoremiftruei 4091 Inference associated with iftrue 4090. (Contributed by BJ, 7-Oct-2018.)
𝜑       if(𝜑, 𝐴, 𝐵) = 𝐴

Theoremiftrued 4092 Value of the conditional operator when its first argument is true. (Contributed by Glauco Siliprandi, 11-Dec-2019.)
(𝜑𝜒)       (𝜑 → if(𝜒, 𝐴, 𝐵) = 𝐴)

Theoremiffalse 4093 Value of the conditional operator when its first argument is false. (Contributed by NM, 14-Aug-1999.)
𝜑 → if(𝜑, 𝐴, 𝐵) = 𝐵)

Theoremiffalsei 4094 Inference associated with iffalse 4093. (Contributed by BJ, 7-Oct-2018.)
¬ 𝜑       if(𝜑, 𝐴, 𝐵) = 𝐵

Theoremiffalsed 4095 Value of the conditional operator when its first argument is false. (Contributed by Glauco Siliprandi, 11-Dec-2019.)
(𝜑 → ¬ 𝜒)       (𝜑 → if(𝜒, 𝐴, 𝐵) = 𝐵)

Theoremifnefalse 4096 When values are unequal, but an "if" condition checks if they are equal, then the "false" branch results. This is a simple utility to provide a slight shortening and simplification of proofs vs. applying iffalse 4093 directly in this case. It happens, e.g., in oevn0 7592. (Contributed by David A. Wheeler, 15-May-2015.)
(𝐴𝐵 → if(𝐴 = 𝐵, 𝐶, 𝐷) = 𝐷)

Theoremifsb 4097 Distribute a function over an if-clause. (Contributed by Mario Carneiro, 14-Aug-2013.)
(if(𝜑, 𝐴, 𝐵) = 𝐴𝐶 = 𝐷)    &   (if(𝜑, 𝐴, 𝐵) = 𝐵𝐶 = 𝐸)       𝐶 = if(𝜑, 𝐷, 𝐸)

Theoremdfif3 4098* Alternate definition of the conditional operator df-if 4085. Note that 𝜑 is independent of 𝑥 i.e. a constant true or false. (Contributed by NM, 25-Aug-2013.) (Revised by Mario Carneiro, 8-Sep-2013.)
𝐶 = {𝑥𝜑}       if(𝜑, 𝐴, 𝐵) = ((𝐴𝐶) ∪ (𝐵 ∩ (V ∖ 𝐶)))

Theoremdfif4 4099* Alternate definition of the conditional operator df-if 4085. Note that 𝜑 is independent of 𝑥 i.e. a constant true or false. (Contributed by NM, 25-Aug-2013.)
𝐶 = {𝑥𝜑}       if(𝜑, 𝐴, 𝐵) = ((𝐴𝐵) ∩ ((𝐴 ∪ (V ∖ 𝐶)) ∩ (𝐵𝐶)))

Theoremdfif5 4100* Alternate definition of the conditional operator df-if 4085. Note that 𝜑 is independent of 𝑥 i.e. a constant true or false (see also ab0orv 3951). (Contributed by Gérard Lang, 18-Aug-2013.)
𝐶 = {𝑥𝜑}       if(𝜑, 𝐴, 𝐵) = ((𝐴𝐵) ∪ (((𝐴𝐵) ∩ 𝐶) ∪ ((𝐵𝐴) ∩ (V ∖ 𝐶))))

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 7301-7400 75 7401-7500 76 7501-7600 77 7601-7700 78 7701-7800 79 7801-7900 80 7901-8000 81 8001-8100 82 8101-8200 83 8201-8300 84 8301-8400 85 8401-8500 86 8501-8600 87 8601-8700 88 8701-8800 89 8801-8900 90 8901-9000 91 9001-9100 92 9101-9200 93 9201-9300 94 9301-9400 95 9401-9500 96 9501-9600 97 9601-9700 98 9701-9800 99 9801-9900 100 9901-10000 101 10001-10100 102 10101-10200 103 10201-10300 104 10301-10400 105 10401-10500 106 10501-10600 107 10601-10700 108 10701-10800 109 10801-10900 110 10901-11000 111 11001-11100 112 11101-11200 113 11201-11300 114 11301-11400 115 11401-11500 116 11501-11600 117 11601-11700 118 11701-11800 119 11801-11900 120 11901-12000 121 12001-12100 122 12101-12200 123 12201-12300 124 12301-12400 125 12401-12500 126 12501-12600 127 12601-12700 128 12701-12800 129 12801-12900 130 12901-13000 131 13001-13100 132 13101-13200 133 13201-13300 134 13301-13400 135 13401-13500 136 13501-13600 137 13601-13700 138 13701-13800 139 13801-13900 140 13901-14000 141 14001-14100 142 14101-14200 143 14201-14300 144 14301-14400 145 14401-14500 146 14501-14600 147 14601-14700 148 14701-14800 149 14801-14900 150 14901-15000 151 15001-15100 152 15101-15200 153 15201-15300 154 15301-15400 155 15401-15500 156 15501-15600 157 15601-15700 158 15701-15800 159 15801-15900 160 15901-16000 161 16001-16100 162 16101-16200 163 16201-16300 164 16301-16400 165 16401-16500 166 16501-16600 167 16601-16700 168 16701-16800 169 16801-16900 170 16901-17000 171 17001-17100 172 17101-17200 173 17201-17300 174 17301-17400 175 17401-17500 176 17501-17600 177 17601-17700 178 17701-17800 179 17801-17900 180 17901-18000 181 18001-18100 182 18101-18200 183 18201-18300 184 18301-18400 185 18401-18500 186 18501-18600 187 18601-18700 188 18701-18800 189 18801-18900 190 18901-19000 191 19001-19100 192 19101-19200 193 19201-19300 194 19301-19400 195 19401-19500 196 19501-19600 197 19601-19700 198 19701-19800 199 19801-19900 200 19901-20000 201 20001-20100 202 20101-20200 203 20201-20300 204 20301-20400 205 20401-20500 206 20501-20600 207 20601-20700 208 20701-20800 209 20801-20900 210 20901-21000 211 21001-21100 212 21101-21200 213 21201-21300 214 21301-21400 215 21401-21500 216 21501-21600 217 21601-21700 218 21701-21800 219 21801-21900 220 21901-22000 221 22001-22100 222 22101-22200 223 22201-22300 224 22301-22400 225 22401-22500 226 22501-22600 227 22601-22700 228 22701-22800 229 22801-22900 230 22901-23000 231 23001-23100 232 23101-23200 233 23201-23300 234 23301-23400 235 23401-23500 236 23501-23600 237 23601-23700 238 23701-23800 239 23801-23900 240 23901-24000 241 24001-24100 242 24101-24200 243 24201-24300 244 24301-24400 245 24401-24500 246 24501-24600 247 24601-24700 248 24701-24800 249 24801-24900 250 24901-25000 251 25001-25100 252 25101-25200 253 25201-25300 254 25301-25400 255 25401-25500 256 25501-25600 257 25601-25700 258 25701-25800 259 25801-25900 260 25901-26000 261 26001-26100 262 26101-26200 263 26201-26300 264 26301-26400 265 26401-26500 266 26501-26600 267 26601-26700 268 26701-26800 269 26801-26900 270 26901-27000 271 27001-27100 272 27101-27200 273 27201-27300 274 27301-27400 275 27401-27500 276 27501-27600 277 27601-27700 278 27701-27800 279 27801-27900 280 27901-28000 281 28001-28100 282 28101-28200 283 28201-28300 284 28301-28400 285 28401-28500 286 28501-28600 287 28601-28700 288 28701-28800 289 28801-28900 290 28901-29000 291 29001-29100 292 29101-29200 293 29201-29300 294 29301-29400 295 29401-29500 296 29501-29600 297 29601-29700 298 29701-29800 299 29801-29900 300 29901-30000 301 30001-30100 302 30101-30200 303 30201-30300 304 30301-30400 305 30401-30500 306 30501-30600 307 30601-30700 308 30701-30800 309 30801-30900 310 30901-31000 311 31001-31100 312 31101-31200 313 31201-31300 314 31301-31400 315 31401-31500 316 31501-31600 317 31601-31700 318 31701-31800 319 31801-31900 320 31901-32000 321 32001-32100 322 32101-32200 323 32201-32300 324 32301-32400 325 32401-32500 326 32501-32600 327 32601-32700 328 32701-32800 329 32801-32900 330 32901-33000 331 33001-33100 332 33101-33200 333 33201-33300 334 33301-33400 335 33401-33500 336 33501-33600 337 33601-33700 338 33701-33800 339 33801-33900 340 33901-34000 341 34001-34100 342 34101-34200 343 34201-34300 344 34301-34400 345 34401-34500 346 34501-34600 347 34601-34700 348 34701-34800 349 34801-34900 350 34901-35000 351 35001-35100 352 35101-35200 353 35201-35300 354 35301-35400 355 35401-35500 356 35501-35600 357 35601-35700 358 35701-35800 359 35801-35900 360 35901-36000 361 36001-36100 362 36101-36200 363 36201-36300 364 36301-36400 365 36401-36500 366 36501-36600 367 36601-36700 368 36701-36800 369 36801-36900 370 36901-37000 371 37001-37100 372 37101-37200 373 37201-37300 374 37301-37400 375 37401-37500 376 37501-37600 377 37601-37700 378 37701-37800 379 37801-37900 380 37901-38000 381 38001-38100 382 38101-38200 383 38201-38300 384 38301-38400 385 38401-38500 386 38501-38600 387 38601-38700 388 38701-38800 389 38801-38900 390 38901-39000 391 39001-39100 392 39101-39200 393 39201-39300 394 39301-39400 395 39401-39500 396 39501-39600 397 39601-39700 398 39701-39800 399 39801-39900 400 39901-40000 401 40001-40100 402 40101-40200 403 40201-40300 404 40301-40400 405 40401-40500 406 40501-40600 407 40601-40700 408 40701-40800 409 40801-40900 410 40901-41000 411 41001-41100 412 41101-41200 413 41201-41300 414 41301-41400 415 41401-41500 416 41501-41600 417 41601-41700 418 41701-41800 419 41801-41900 420 41901-42000 421 42001-42100 422 42101-42200 423 42201-42300 424 42301-42322
 Copyright terms: Public domain < Previous  Next >