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Theorem List for Metamath Proof Explorer - 35401-35500   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremisidlc 35401* The predicate "is an ideal of the commutative ring 𝑅". (Contributed by Jeff Madsen, 10-Jun-2010.)
𝐺 = (1st𝑅)    &   𝐻 = (2nd𝑅)    &   𝑋 = ran 𝐺    &   𝑍 = (GId‘𝐺)       (𝑅 ∈ CRingOps → (𝐼 ∈ (Idl‘𝑅) ↔ (𝐼𝑋𝑍𝐼 ∧ ∀𝑥𝐼 (∀𝑦𝐼 (𝑥𝐺𝑦) ∈ 𝐼 ∧ ∀𝑧𝑋 (𝑧𝐻𝑥) ∈ 𝐼))))

Theoremidlss 35402 An ideal of 𝑅 is a subset of 𝑅. (Contributed by Jeff Madsen, 10-Jun-2010.)
𝐺 = (1st𝑅)    &   𝑋 = ran 𝐺       ((𝑅 ∈ RingOps ∧ 𝐼 ∈ (Idl‘𝑅)) → 𝐼𝑋)

Theoremidlcl 35403 An element of an ideal is an element of the ring. (Contributed by Jeff Madsen, 19-Jun-2010.)
𝐺 = (1st𝑅)    &   𝑋 = ran 𝐺       (((𝑅 ∈ RingOps ∧ 𝐼 ∈ (Idl‘𝑅)) ∧ 𝐴𝐼) → 𝐴𝑋)

Theoremidl0cl 35404 An ideal contains 0. (Contributed by Jeff Madsen, 10-Jun-2010.)
𝐺 = (1st𝑅)    &   𝑍 = (GId‘𝐺)       ((𝑅 ∈ RingOps ∧ 𝐼 ∈ (Idl‘𝑅)) → 𝑍𝐼)

𝐺 = (1st𝑅)       (((𝑅 ∈ RingOps ∧ 𝐼 ∈ (Idl‘𝑅)) ∧ (𝐴𝐼𝐵𝐼)) → (𝐴𝐺𝐵) ∈ 𝐼)

Theoremidllmulcl 35406 An ideal is closed under multiplication on the left. (Contributed by Jeff Madsen, 10-Jun-2010.)
𝐺 = (1st𝑅)    &   𝐻 = (2nd𝑅)    &   𝑋 = ran 𝐺       (((𝑅 ∈ RingOps ∧ 𝐼 ∈ (Idl‘𝑅)) ∧ (𝐴𝐼𝐵𝑋)) → (𝐵𝐻𝐴) ∈ 𝐼)

Theoremidlrmulcl 35407 An ideal is closed under multiplication on the right. (Contributed by Jeff Madsen, 10-Jun-2010.)
𝐺 = (1st𝑅)    &   𝐻 = (2nd𝑅)    &   𝑋 = ran 𝐺       (((𝑅 ∈ RingOps ∧ 𝐼 ∈ (Idl‘𝑅)) ∧ (𝐴𝐼𝐵𝑋)) → (𝐴𝐻𝐵) ∈ 𝐼)

Theoremidlnegcl 35408 An ideal is closed under negation. (Contributed by Jeff Madsen, 10-Jun-2010.)
𝐺 = (1st𝑅)    &   𝑁 = (inv‘𝐺)       (((𝑅 ∈ RingOps ∧ 𝐼 ∈ (Idl‘𝑅)) ∧ 𝐴𝐼) → (𝑁𝐴) ∈ 𝐼)

Theoremidlsubcl 35409 An ideal is closed under subtraction. (Contributed by Jeff Madsen, 19-Jun-2010.)
𝐺 = (1st𝑅)    &   𝐷 = ( /𝑔𝐺)       (((𝑅 ∈ RingOps ∧ 𝐼 ∈ (Idl‘𝑅)) ∧ (𝐴𝐼𝐵𝐼)) → (𝐴𝐷𝐵) ∈ 𝐼)

Theoremrngoidl 35410 A ring 𝑅 is an 𝑅 ideal. (Contributed by Jeff Madsen, 10-Jun-2010.)
𝐺 = (1st𝑅)    &   𝑋 = ran 𝐺       (𝑅 ∈ RingOps → 𝑋 ∈ (Idl‘𝑅))

Theorem0idl 35411 The set containing only 0 is an ideal. (Contributed by Jeff Madsen, 10-Jun-2010.)
𝐺 = (1st𝑅)    &   𝑍 = (GId‘𝐺)       (𝑅 ∈ RingOps → {𝑍} ∈ (Idl‘𝑅))

Theorem1idl 35412 Two ways of expressing the unit ideal. (Contributed by Jeff Madsen, 10-Jun-2010.)
𝐺 = (1st𝑅)    &   𝐻 = (2nd𝑅)    &   𝑋 = ran 𝐺    &   𝑈 = (GId‘𝐻)       ((𝑅 ∈ RingOps ∧ 𝐼 ∈ (Idl‘𝑅)) → (𝑈𝐼𝐼 = 𝑋))

Theorem0rngo 35413 In a ring, 0 = 1 iff the ring contains only 0. (Contributed by Jeff Madsen, 6-Jan-2011.)
𝐺 = (1st𝑅)    &   𝐻 = (2nd𝑅)    &   𝑋 = ran 𝐺    &   𝑍 = (GId‘𝐺)    &   𝑈 = (GId‘𝐻)       (𝑅 ∈ RingOps → (𝑍 = 𝑈𝑋 = {𝑍}))

Theoremdivrngidl 35414 The only ideals in a division ring are the zero ideal and the unit ideal. (Contributed by Jeff Madsen, 10-Jun-2010.)
𝐺 = (1st𝑅)    &   𝐻 = (2nd𝑅)    &   𝑋 = ran 𝐺    &   𝑍 = (GId‘𝐺)       (𝑅 ∈ DivRingOps → (Idl‘𝑅) = {{𝑍}, 𝑋})

Theoremintidl 35415 The intersection of a nonempty collection of ideals is an ideal. (Contributed by Jeff Madsen, 10-Jun-2010.)
((𝑅 ∈ RingOps ∧ 𝐶 ≠ ∅ ∧ 𝐶 ⊆ (Idl‘𝑅)) → 𝐶 ∈ (Idl‘𝑅))

Theoreminidl 35416 The intersection of two ideals is an ideal. (Contributed by Jeff Madsen, 16-Jun-2011.)
((𝑅 ∈ RingOps ∧ 𝐼 ∈ (Idl‘𝑅) ∧ 𝐽 ∈ (Idl‘𝑅)) → (𝐼𝐽) ∈ (Idl‘𝑅))

Theoremunichnidl 35417* The union of a nonempty chain of ideals is an ideal. (Contributed by Jeff Madsen, 5-Jan-2011.)
((𝑅 ∈ RingOps ∧ (𝐶 ≠ ∅ ∧ 𝐶 ⊆ (Idl‘𝑅) ∧ ∀𝑖𝐶𝑗𝐶 (𝑖𝑗𝑗𝑖))) → 𝐶 ∈ (Idl‘𝑅))

Theoremkeridl 35418 The kernel of a ring homomorphism is an ideal. (Contributed by Jeff Madsen, 3-Jan-2011.)
𝐺 = (1st𝑆)    &   𝑍 = (GId‘𝐺)       ((𝑅 ∈ RingOps ∧ 𝑆 ∈ RingOps ∧ 𝐹 ∈ (𝑅 RngHom 𝑆)) → (𝐹 “ {𝑍}) ∈ (Idl‘𝑅))

Theorempridlval 35419* The class of prime ideals of a ring 𝑅. (Contributed by Jeff Madsen, 10-Jun-2010.)
𝐺 = (1st𝑅)    &   𝐻 = (2nd𝑅)    &   𝑋 = ran 𝐺       (𝑅 ∈ RingOps → (PrIdl‘𝑅) = {𝑖 ∈ (Idl‘𝑅) ∣ (𝑖𝑋 ∧ ∀𝑎 ∈ (Idl‘𝑅)∀𝑏 ∈ (Idl‘𝑅)(∀𝑥𝑎𝑦𝑏 (𝑥𝐻𝑦) ∈ 𝑖 → (𝑎𝑖𝑏𝑖)))})

Theoremispridl 35420* The predicate "is a prime ideal". (Contributed by Jeff Madsen, 10-Jun-2010.)
𝐺 = (1st𝑅)    &   𝐻 = (2nd𝑅)    &   𝑋 = ran 𝐺       (𝑅 ∈ RingOps → (𝑃 ∈ (PrIdl‘𝑅) ↔ (𝑃 ∈ (Idl‘𝑅) ∧ 𝑃𝑋 ∧ ∀𝑎 ∈ (Idl‘𝑅)∀𝑏 ∈ (Idl‘𝑅)(∀𝑥𝑎𝑦𝑏 (𝑥𝐻𝑦) ∈ 𝑃 → (𝑎𝑃𝑏𝑃)))))

Theorempridlidl 35421 A prime ideal is an ideal. (Contributed by Jeff Madsen, 19-Jun-2010.)
((𝑅 ∈ RingOps ∧ 𝑃 ∈ (PrIdl‘𝑅)) → 𝑃 ∈ (Idl‘𝑅))

Theorempridlnr 35422 A prime ideal is a proper ideal. (Contributed by Jeff Madsen, 19-Jun-2010.)
𝐺 = (1st𝑅)    &   𝑋 = ran 𝐺       ((𝑅 ∈ RingOps ∧ 𝑃 ∈ (PrIdl‘𝑅)) → 𝑃𝑋)

Theorempridl 35423* The main property of a prime ideal. (Contributed by Jeff Madsen, 19-Jun-2010.)
𝐻 = (2nd𝑅)       (((𝑅 ∈ RingOps ∧ 𝑃 ∈ (PrIdl‘𝑅)) ∧ (𝐴 ∈ (Idl‘𝑅) ∧ 𝐵 ∈ (Idl‘𝑅) ∧ ∀𝑥𝐴𝑦𝐵 (𝑥𝐻𝑦) ∈ 𝑃)) → (𝐴𝑃𝐵𝑃))

Theoremispridl2 35424* A condition that shows an ideal is prime. For commutative rings, this is often taken to be the definition. See ispridlc 35456 for the equivalence in the commutative case. (Contributed by Jeff Madsen, 19-Jun-2010.)
𝐺 = (1st𝑅)    &   𝐻 = (2nd𝑅)    &   𝑋 = ran 𝐺       ((𝑅 ∈ RingOps ∧ (𝑃 ∈ (Idl‘𝑅) ∧ 𝑃𝑋 ∧ ∀𝑎𝑋𝑏𝑋 ((𝑎𝐻𝑏) ∈ 𝑃 → (𝑎𝑃𝑏𝑃)))) → 𝑃 ∈ (PrIdl‘𝑅))

Theoremmaxidlval 35425* The set of maximal ideals of a ring. (Contributed by Jeff Madsen, 5-Jan-2011.)
𝐺 = (1st𝑅)    &   𝑋 = ran 𝐺       (𝑅 ∈ RingOps → (MaxIdl‘𝑅) = {𝑖 ∈ (Idl‘𝑅) ∣ (𝑖𝑋 ∧ ∀𝑗 ∈ (Idl‘𝑅)(𝑖𝑗 → (𝑗 = 𝑖𝑗 = 𝑋)))})

Theoremismaxidl 35426* The predicate "is a maximal ideal". (Contributed by Jeff Madsen, 5-Jan-2011.)
𝐺 = (1st𝑅)    &   𝑋 = ran 𝐺       (𝑅 ∈ RingOps → (𝑀 ∈ (MaxIdl‘𝑅) ↔ (𝑀 ∈ (Idl‘𝑅) ∧ 𝑀𝑋 ∧ ∀𝑗 ∈ (Idl‘𝑅)(𝑀𝑗 → (𝑗 = 𝑀𝑗 = 𝑋)))))

Theoremmaxidlidl 35427 A maximal ideal is an ideal. (Contributed by Jeff Madsen, 5-Jan-2011.)
((𝑅 ∈ RingOps ∧ 𝑀 ∈ (MaxIdl‘𝑅)) → 𝑀 ∈ (Idl‘𝑅))

Theoremmaxidlnr 35428 A maximal ideal is proper. (Contributed by Jeff Madsen, 16-Jun-2011.)
𝐺 = (1st𝑅)    &   𝑋 = ran 𝐺       ((𝑅 ∈ RingOps ∧ 𝑀 ∈ (MaxIdl‘𝑅)) → 𝑀𝑋)

Theoremmaxidlmax 35429 A maximal ideal is a maximal proper ideal. (Contributed by Jeff Madsen, 16-Jun-2011.)
𝐺 = (1st𝑅)    &   𝑋 = ran 𝐺       (((𝑅 ∈ RingOps ∧ 𝑀 ∈ (MaxIdl‘𝑅)) ∧ (𝐼 ∈ (Idl‘𝑅) ∧ 𝑀𝐼)) → (𝐼 = 𝑀𝐼 = 𝑋))

Theoremmaxidln1 35430 One is not contained in any maximal ideal. (Contributed by Jeff Madsen, 17-Jun-2011.)
𝐻 = (2nd𝑅)    &   𝑈 = (GId‘𝐻)       ((𝑅 ∈ RingOps ∧ 𝑀 ∈ (MaxIdl‘𝑅)) → ¬ 𝑈𝑀)

Theoremmaxidln0 35431 A ring with a maximal ideal is not the zero ring. (Contributed by Jeff Madsen, 17-Jun-2011.)
𝐺 = (1st𝑅)    &   𝐻 = (2nd𝑅)    &   𝑍 = (GId‘𝐺)    &   𝑈 = (GId‘𝐻)       ((𝑅 ∈ RingOps ∧ 𝑀 ∈ (MaxIdl‘𝑅)) → 𝑈𝑍)

20.20.21  Prime rings and integral domains

Syntaxcprrng 35432 Extend class notation with the class of prime rings.
class PrRing

Syntaxcdmn 35433 Extend class notation with the class of domains.
class Dmn

Definitiondf-prrngo 35434 Define the class of prime rings. A ring is prime if the zero ideal is a prime ideal. (Contributed by Jeff Madsen, 10-Jun-2010.)
PrRing = {𝑟 ∈ RingOps ∣ {(GId‘(1st𝑟))} ∈ (PrIdl‘𝑟)}

Definitiondf-dmn 35435 Define the class of (integral) domains. A domain is a commutative prime ring. (Contributed by Jeff Madsen, 10-Jun-2010.)
Dmn = (PrRing ∩ Com2)

Theoremisprrngo 35436 The predicate "is a prime ring". (Contributed by Jeff Madsen, 10-Jun-2010.)
𝐺 = (1st𝑅)    &   𝑍 = (GId‘𝐺)       (𝑅 ∈ PrRing ↔ (𝑅 ∈ RingOps ∧ {𝑍} ∈ (PrIdl‘𝑅)))

Theoremprrngorngo 35437 A prime ring is a ring. (Contributed by Jeff Madsen, 10-Jun-2010.)
(𝑅 ∈ PrRing → 𝑅 ∈ RingOps)

Theoremsmprngopr 35438 A simple ring (one whose only ideals are 0 and 𝑅) is a prime ring. (Contributed by Jeff Madsen, 6-Jan-2011.)
𝐺 = (1st𝑅)    &   𝐻 = (2nd𝑅)    &   𝑋 = ran 𝐺    &   𝑍 = (GId‘𝐺)    &   𝑈 = (GId‘𝐻)       ((𝑅 ∈ RingOps ∧ 𝑈𝑍 ∧ (Idl‘𝑅) = {{𝑍}, 𝑋}) → 𝑅 ∈ PrRing)

Theoremdivrngpr 35439 A division ring is a prime ring. (Contributed by Jeff Madsen, 6-Jan-2011.)
(𝑅 ∈ DivRingOps → 𝑅 ∈ PrRing)

Theoremisdmn 35440 The predicate "is a domain". (Contributed by Jeff Madsen, 10-Jun-2010.)
(𝑅 ∈ Dmn ↔ (𝑅 ∈ PrRing ∧ 𝑅 ∈ Com2))

Theoremisdmn2 35441 The predicate "is a domain". (Contributed by Jeff Madsen, 10-Jun-2010.)
(𝑅 ∈ Dmn ↔ (𝑅 ∈ PrRing ∧ 𝑅 ∈ CRingOps))

Theoremdmncrng 35442 A domain is a commutative ring. (Contributed by Jeff Madsen, 6-Jan-2011.)
(𝑅 ∈ Dmn → 𝑅 ∈ CRingOps)

Theoremdmnrngo 35443 A domain is a ring. (Contributed by Jeff Madsen, 6-Jan-2011.)
(𝑅 ∈ Dmn → 𝑅 ∈ RingOps)

Theoremflddmn 35444 A field is a domain. (Contributed by Jeff Madsen, 10-Jun-2010.)
(𝐾 ∈ Fld → 𝐾 ∈ Dmn)

20.20.22  Ideal generators

Syntaxcigen 35445 Extend class notation with the ideal generation function.
class IdlGen

Definitiondf-igen 35446* Define the ideal generated by a subset of a ring. (Contributed by Jeff Madsen, 10-Jun-2010.)
IdlGen = (𝑟 ∈ RingOps, 𝑠 ∈ 𝒫 ran (1st𝑟) ↦ {𝑗 ∈ (Idl‘𝑟) ∣ 𝑠𝑗})

Theoremigenval 35447* The ideal generated by a subset of a ring. (Contributed by Jeff Madsen, 10-Jun-2010.) (Proof shortened by Mario Carneiro, 20-Dec-2013.)
𝐺 = (1st𝑅)    &   𝑋 = ran 𝐺       ((𝑅 ∈ RingOps ∧ 𝑆𝑋) → (𝑅 IdlGen 𝑆) = {𝑗 ∈ (Idl‘𝑅) ∣ 𝑆𝑗})

Theoremigenss 35448 A set is a subset of the ideal it generates. (Contributed by Jeff Madsen, 10-Jun-2010.)
𝐺 = (1st𝑅)    &   𝑋 = ran 𝐺       ((𝑅 ∈ RingOps ∧ 𝑆𝑋) → 𝑆 ⊆ (𝑅 IdlGen 𝑆))

Theoremigenidl 35449 The ideal generated by a set is an ideal. (Contributed by Jeff Madsen, 10-Jun-2010.)
𝐺 = (1st𝑅)    &   𝑋 = ran 𝐺       ((𝑅 ∈ RingOps ∧ 𝑆𝑋) → (𝑅 IdlGen 𝑆) ∈ (Idl‘𝑅))

Theoremigenmin 35450 The ideal generated by a set is the minimal ideal containing that set. (Contributed by Jeff Madsen, 10-Jun-2010.)
((𝑅 ∈ RingOps ∧ 𝐼 ∈ (Idl‘𝑅) ∧ 𝑆𝐼) → (𝑅 IdlGen 𝑆) ⊆ 𝐼)

Theoremigenidl2 35451 The ideal generated by an ideal is that ideal. (Contributed by Jeff Madsen, 10-Jun-2010.)
((𝑅 ∈ RingOps ∧ 𝐼 ∈ (Idl‘𝑅)) → (𝑅 IdlGen 𝐼) = 𝐼)

Theoremigenval2 35452* The ideal generated by a subset of a ring. (Contributed by Jeff Madsen, 10-Jun-2010.)
𝐺 = (1st𝑅)    &   𝑋 = ran 𝐺       ((𝑅 ∈ RingOps ∧ 𝑆𝑋) → ((𝑅 IdlGen 𝑆) = 𝐼 ↔ (𝐼 ∈ (Idl‘𝑅) ∧ 𝑆𝐼 ∧ ∀𝑗 ∈ (Idl‘𝑅)(𝑆𝑗𝐼𝑗))))

Theoremprnc 35453* A principal ideal (an ideal generated by one element) in a commutative ring. (Contributed by Jeff Madsen, 10-Jun-2010.)
𝐺 = (1st𝑅)    &   𝐻 = (2nd𝑅)    &   𝑋 = ran 𝐺       ((𝑅 ∈ CRingOps ∧ 𝐴𝑋) → (𝑅 IdlGen {𝐴}) = {𝑥𝑋 ∣ ∃𝑦𝑋 𝑥 = (𝑦𝐻𝐴)})

Theoremisfldidl 35454 Determine if a ring is a field based on its ideals. (Contributed by Jeff Madsen, 10-Jun-2010.)
𝐺 = (1st𝐾)    &   𝐻 = (2nd𝐾)    &   𝑋 = ran 𝐺    &   𝑍 = (GId‘𝐺)    &   𝑈 = (GId‘𝐻)       (𝐾 ∈ Fld ↔ (𝐾 ∈ CRingOps ∧ 𝑈𝑍 ∧ (Idl‘𝐾) = {{𝑍}, 𝑋}))

Theoremisfldidl2 35455 Determine if a ring is a field based on its ideals. (Contributed by Jeff Madsen, 6-Jan-2011.)
𝐺 = (1st𝐾)    &   𝐻 = (2nd𝐾)    &   𝑋 = ran 𝐺    &   𝑍 = (GId‘𝐺)       (𝐾 ∈ Fld ↔ (𝐾 ∈ CRingOps ∧ 𝑋 ≠ {𝑍} ∧ (Idl‘𝐾) = {{𝑍}, 𝑋}))

Theoremispridlc 35456* The predicate "is a prime ideal". Alternate definition for commutative rings. (Contributed by Jeff Madsen, 19-Jun-2010.)
𝐺 = (1st𝑅)    &   𝐻 = (2nd𝑅)    &   𝑋 = ran 𝐺       (𝑅 ∈ CRingOps → (𝑃 ∈ (PrIdl‘𝑅) ↔ (𝑃 ∈ (Idl‘𝑅) ∧ 𝑃𝑋 ∧ ∀𝑎𝑋𝑏𝑋 ((𝑎𝐻𝑏) ∈ 𝑃 → (𝑎𝑃𝑏𝑃)))))

Theorempridlc 35457 Property of a prime ideal in a commutative ring. (Contributed by Jeff Madsen, 17-Jun-2011.)
𝐺 = (1st𝑅)    &   𝐻 = (2nd𝑅)    &   𝑋 = ran 𝐺       (((𝑅 ∈ CRingOps ∧ 𝑃 ∈ (PrIdl‘𝑅)) ∧ (𝐴𝑋𝐵𝑋 ∧ (𝐴𝐻𝐵) ∈ 𝑃)) → (𝐴𝑃𝐵𝑃))

Theorempridlc2 35458 Property of a prime ideal in a commutative ring. (Contributed by Jeff Madsen, 17-Jun-2011.)
𝐺 = (1st𝑅)    &   𝐻 = (2nd𝑅)    &   𝑋 = ran 𝐺       (((𝑅 ∈ CRingOps ∧ 𝑃 ∈ (PrIdl‘𝑅)) ∧ (𝐴 ∈ (𝑋𝑃) ∧ 𝐵𝑋 ∧ (𝐴𝐻𝐵) ∈ 𝑃)) → 𝐵𝑃)

Theorempridlc3 35459 Property of a prime ideal in a commutative ring. (Contributed by Jeff Madsen, 17-Jun-2011.)
𝐺 = (1st𝑅)    &   𝐻 = (2nd𝑅)    &   𝑋 = ran 𝐺       (((𝑅 ∈ CRingOps ∧ 𝑃 ∈ (PrIdl‘𝑅)) ∧ (𝐴 ∈ (𝑋𝑃) ∧ 𝐵 ∈ (𝑋𝑃))) → (𝐴𝐻𝐵) ∈ (𝑋𝑃))

Theoremisdmn3 35460* The predicate "is a domain", alternate expression. (Contributed by Jeff Madsen, 19-Jun-2010.)
𝐺 = (1st𝑅)    &   𝐻 = (2nd𝑅)    &   𝑋 = ran 𝐺    &   𝑍 = (GId‘𝐺)    &   𝑈 = (GId‘𝐻)       (𝑅 ∈ Dmn ↔ (𝑅 ∈ CRingOps ∧ 𝑈𝑍 ∧ ∀𝑎𝑋𝑏𝑋 ((𝑎𝐻𝑏) = 𝑍 → (𝑎 = 𝑍𝑏 = 𝑍))))

Theoremdmnnzd 35461 A domain has no zero-divisors (besides zero). (Contributed by Jeff Madsen, 19-Jun-2010.)
𝐺 = (1st𝑅)    &   𝐻 = (2nd𝑅)    &   𝑋 = ran 𝐺    &   𝑍 = (GId‘𝐺)       ((𝑅 ∈ Dmn ∧ (𝐴𝑋𝐵𝑋 ∧ (𝐴𝐻𝐵) = 𝑍)) → (𝐴 = 𝑍𝐵 = 𝑍))

Theoremdmncan1 35462 Cancellation law for domains. (Contributed by Jeff Madsen, 6-Jan-2011.)
𝐺 = (1st𝑅)    &   𝐻 = (2nd𝑅)    &   𝑋 = ran 𝐺    &   𝑍 = (GId‘𝐺)       (((𝑅 ∈ Dmn ∧ (𝐴𝑋𝐵𝑋𝐶𝑋)) ∧ 𝐴𝑍) → ((𝐴𝐻𝐵) = (𝐴𝐻𝐶) → 𝐵 = 𝐶))

Theoremdmncan2 35463 Cancellation law for domains. (Contributed by Jeff Madsen, 6-Jan-2011.)
𝐺 = (1st𝑅)    &   𝐻 = (2nd𝑅)    &   𝑋 = ran 𝐺    &   𝑍 = (GId‘𝐺)       (((𝑅 ∈ Dmn ∧ (𝐴𝑋𝐵𝑋𝐶𝑋)) ∧ 𝐶𝑍) → ((𝐴𝐻𝐶) = (𝐵𝐻𝐶) → 𝐴 = 𝐵))

20.21  Mathbox for Giovanni Mascellani

20.21.1  Tools for automatic proof building

The results in this section are mostly meant for being used by automatic proof building programs. As a result, they might appear less useful or meaningful than others to human beings.

Theoremefald2 35464 A proof by contradiction. (Contributed by Giovanni Mascellani, 15-Sep-2017.)
𝜑 → ⊥)       𝜑

Theoremnotbinot1 35465 Simplification rule of negation across a biimplication. (Contributed by Giovanni Mascellani, 15-Sep-2017.)
(¬ (¬ 𝜑𝜓) ↔ (𝜑𝜓))

Theorembicontr 35466 Biimplication of its own negation is a contradiction. (Contributed by Giovanni Mascellani, 15-Sep-2017.)
((¬ 𝜑𝜑) ↔ ⊥)

Theoremimpor 35467 An equivalent formula for implying a disjunction. (Contributed by Giovanni Mascellani, 15-Sep-2017.)
((𝜑 → (𝜓𝜒)) ↔ ((¬ 𝜑𝜓) ∨ 𝜒))

Theoremorfa 35468 The falsum can be removed from a disjunction. (Contributed by Giovanni Mascellani, 15-Sep-2017.)
((𝜑 ∨ ⊥) ↔ 𝜑)

Theoremnotbinot2 35469 Commutation rule between negation and biimplication. (Contributed by Giovanni Mascellani, 15-Sep-2017.)
(¬ (𝜑𝜓) ↔ (¬ 𝜑𝜓))

Theorembiimpor 35470 A rewriting rule for biimplication. (Contributed by Giovanni Mascellani, 15-Sep-2017.)
(((𝜑𝜓) → 𝜒) ↔ ((¬ 𝜑𝜓) ∨ 𝜒))

Theoremorfa1 35471 Add a contradicting disjunct to an antecedent. (Contributed by Giovanni Mascellani, 15-Sep-2017.)
(𝜑𝜓)       ((𝜑 ∨ ⊥) → 𝜓)

Theoremorfa2 35472 Remove a contradicting disjunct from an antecedent. (Contributed by Giovanni Mascellani, 15-Sep-2017.)
(𝜑 → ⊥)       ((𝜑𝜓) → 𝜓)

Theorembifald 35473 Infer the equivalence to a contradiction from a negation, in deduction form. (Contributed by Giovanni Mascellani, 15-Sep-2017.)
(𝜑 → ¬ 𝜓)       (𝜑 → (𝜓 ↔ ⊥))

Theoremorsild 35474 A lemma for not-or-not elimination, in deduction form. (Contributed by Giovanni Mascellani, 15-Sep-2017.)
(𝜑 → ¬ (𝜓𝜒))       (𝜑 → ¬ 𝜓)

Theoremorsird 35475 A lemma for not-or-not elimination, in deduction form. (Contributed by Giovanni Mascellani, 15-Sep-2017.)
(𝜑 → ¬ (𝜓𝜒))       (𝜑 → ¬ 𝜒)

Theoremcnf1dd 35476 A lemma for Conjunctive Normal Form unit propagation, in double deduction form. (Contributed by Giovanni Mascellani, 19-Mar-2018.)
(𝜑 → (𝜓 → ¬ 𝜒))    &   (𝜑 → (𝜓 → (𝜒𝜃)))       (𝜑 → (𝜓𝜃))

Theoremcnf2dd 35477 A lemma for Conjunctive Normal Form unit propagation, in double deduction form. (Contributed by Giovanni Mascellani, 19-Mar-2018.)
(𝜑 → (𝜓 → ¬ 𝜃))    &   (𝜑 → (𝜓 → (𝜒𝜃)))       (𝜑 → (𝜓𝜒))

Theoremcnfn1dd 35478 A lemma for Conjunctive Normal Form unit propagation, in double deduction form. (Contributed by Giovanni Mascellani, 19-Mar-2018.)
(𝜑 → (𝜓𝜒))    &   (𝜑 → (𝜓 → (¬ 𝜒𝜃)))       (𝜑 → (𝜓𝜃))

Theoremcnfn2dd 35479 A lemma for Conjunctive Normal Form unit propagation, in double deduction form. (Contributed by Giovanni Mascellani, 19-Mar-2018.)
(𝜑 → (𝜓𝜃))    &   (𝜑 → (𝜓 → (𝜒 ∨ ¬ 𝜃)))       (𝜑 → (𝜓𝜒))

Theoremor32dd 35480 A rearrangement of disjuncts, in double deduction form. (Contributed by Giovanni Mascellani, 19-Mar-2018.)
(𝜑 → (𝜓 → ((𝜒𝜃) ∨ 𝜏)))       (𝜑 → (𝜓 → ((𝜒𝜏) ∨ 𝜃)))

Theoremnotornotel1 35481 A lemma for not-or-not elimination, in deduction form. (Contributed by Giovanni Mascellani, 19-Mar-2018.)
(𝜑 → ¬ (¬ 𝜓𝜒))       (𝜑𝜓)

Theoremnotornotel2 35482 A lemma for not-or-not elimination, in deduction form. (Contributed by Giovanni Mascellani, 19-Mar-2018.)
(𝜑 → ¬ (𝜓 ∨ ¬ 𝜒))       (𝜑𝜒)

Theoremcontrd 35483 A proof by contradiction, in deduction form. (Contributed by Giovanni Mascellani, 19-Mar-2018.)
(𝜑 → (¬ 𝜓𝜒))    &   (𝜑 → (¬ 𝜓 → ¬ 𝜒))       (𝜑𝜓)

Theoreman12i 35484 An inference from commuting operands in a chain of conjunctions. (Contributed by Giovanni Mascellani, 22-May-2019.)
(𝜑 ∧ (𝜓𝜒))       (𝜓 ∧ (𝜑𝜒))

Theoremexmid2 35485 An excluded middle law. (Contributed by Giovanni Mascellani, 23-May-2019.)
((𝜓𝜑) → 𝜒)    &   ((¬ 𝜓𝜂) → 𝜒)       ((𝜑𝜂) → 𝜒)

Theoremselconj 35486 An inference for selecting one of a list of conjuncts. (Contributed by Giovanni Mascellani, 23-May-2019.)
(𝜑 ↔ (𝜓𝜒))       ((𝜂𝜑) ↔ (𝜓 ∧ (𝜂𝜒)))

Theoremtruconj 35487 Add true as a conjunct. (Contributed by Giovanni Mascellani, 23-May-2019.)
(𝜑 ↔ (⊤ ∧ 𝜑))

Theoremorel 35488 An inference for disjunction elimination. (Contributed by Giovanni Mascellani, 24-May-2019.)
((𝜓𝜂) → 𝜃)    &   ((𝜒𝜌) → 𝜃)    &   (𝜑 → (𝜓𝜒))       ((𝜑 ∧ (𝜂𝜌)) → 𝜃)

Theoremnegel 35489 An inference for negation elimination. (Contributed by Giovanni Mascellani, 24-May-2019.)
(𝜓𝜒)    &   (𝜑 → ¬ 𝜒)       ((𝜑𝜓) → ⊥)

Theorembotel 35490 An inference for bottom elimination. (Contributed by Giovanni Mascellani, 24-May-2019.)
(𝜑 → ⊥)       (𝜑𝜓)

(𝜑𝜓)       (𝜑 ↔ (⊤ ∧ 𝜓))

Theoremgm-sbtru 35492 Substitution does not change truth. (Contributed by Giovanni Mascellani, 24-May-2019.)
𝐴 ∈ V       ([𝐴 / 𝑥]⊤ ↔ ⊤)

Theoremsbfal 35493 Substitution does not change falsity. (Contributed by Giovanni Mascellani, 24-May-2019.)
𝐴 ∈ V       ([𝐴 / 𝑥]⊥ ↔ ⊥)

Theoremsbcani 35494 Distribution of class substitution over conjunction, in inference form. (Contributed by Giovanni Mascellani, 27-May-2019.)
([𝐴 / 𝑥]𝜑𝜒)    &   ([𝐴 / 𝑥]𝜓𝜂)       ([𝐴 / 𝑥](𝜑𝜓) ↔ (𝜒𝜂))

Theoremsbcori 35495 Distribution of class substitution over disjunction, in inference form. (Contributed by Giovanni Mascellani, 27-May-2019.)
([𝐴 / 𝑥]𝜑𝜒)    &   ([𝐴 / 𝑥]𝜓𝜂)       ([𝐴 / 𝑥](𝜑𝜓) ↔ (𝜒𝜂))

Theoremsbcimi 35496 Distribution of class substitution over implication, in inference form. (Contributed by Giovanni Mascellani, 27-May-2019.)
𝐴 ∈ V    &   ([𝐴 / 𝑥]𝜑𝜒)    &   ([𝐴 / 𝑥]𝜓𝜂)       ([𝐴 / 𝑥](𝜑𝜓) ↔ (𝜒𝜂))

Theoremsbcni 35497 Move class substitution inside a negation, in inference form. (Contributed by Giovanni Mascellani, 27-May-2019.)
𝐴 ∈ V    &   ([𝐴 / 𝑥]𝜑𝜓)       ([𝐴 / 𝑥] ¬ 𝜑 ↔ ¬ 𝜓)

Theoremsbali 35498 Discard class substitution in a universal quantification when substituting the quantified variable, in inference form. (Contributed by Giovanni Mascellani, 27-May-2019.)
𝐴 ∈ V       ([𝐴 / 𝑥]𝑥𝜑 ↔ ∀𝑥𝜑)

Theoremsbexi 35499 Discard class substitution in an existential quantification when substituting the quantified variable, in inference form. (Contributed by Giovanni Mascellani, 27-May-2019.)
𝐴 ∈ V       ([𝐴 / 𝑥]𝑥𝜑 ↔ ∃𝑥𝜑)

Theoremsbcalf 35500* Move universal quantifier in and out of class substitution, with an explicit non-free variable condition. (Contributed by Giovanni Mascellani, 29-May-2019.)
𝑦𝐴       ([𝐴 / 𝑥]𝑦𝜑 ↔ ∀𝑦[𝐴 / 𝑥]𝜑)

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