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Theorem List for Metamath Proof Explorer - 1401-1500   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremsyld3an3 1401 A syllogism inference. (Contributed by NM, 20-May-2007.)
((𝜑𝜓𝜒) → 𝜃)    &   ((𝜑𝜓𝜃) → 𝜏)       ((𝜑𝜓𝜒) → 𝜏)
 
Theoremsyld3an1 1402 A syllogism inference. (Contributed by NM, 7-Jul-2008.) (Proof shortened by Wolf Lammen, 26-Jun-2022.)
((𝜒𝜓𝜃) → 𝜑)    &   ((𝜑𝜓𝜃) → 𝜏)       ((𝜒𝜓𝜃) → 𝜏)
 
Theoremsyld3an2 1403 A syllogism inference. (Contributed by NM, 20-May-2007.)
((𝜑𝜒𝜃) → 𝜓)    &   ((𝜑𝜓𝜃) → 𝜏)       ((𝜑𝜒𝜃) → 𝜏)
 
Theoremsyl3anl1 1404 A syllogism inference. (Contributed by NM, 24-Feb-2005.)
(𝜑𝜓)    &   (((𝜓𝜒𝜃) ∧ 𝜏) → 𝜂)       (((𝜑𝜒𝜃) ∧ 𝜏) → 𝜂)
 
Theoremsyl3anl2 1405 A syllogism inference. (Contributed by NM, 24-Feb-2005.) (Proof shortened by Wolf Lammen, 27-Jun-2022.)
(𝜑𝜒)    &   (((𝜓𝜒𝜃) ∧ 𝜏) → 𝜂)       (((𝜓𝜑𝜃) ∧ 𝜏) → 𝜂)
 
Theoremsyl3anl3 1406 A syllogism inference. (Contributed by NM, 24-Feb-2005.)
(𝜑𝜃)    &   (((𝜓𝜒𝜃) ∧ 𝜏) → 𝜂)       (((𝜓𝜒𝜑) ∧ 𝜏) → 𝜂)
 
Theoremsyl3anl 1407 A triple syllogism inference. (Contributed by NM, 24-Dec-2006.)
(𝜑𝜓)    &   (𝜒𝜃)    &   (𝜏𝜂)    &   (((𝜓𝜃𝜂) ∧ 𝜁) → 𝜎)       (((𝜑𝜒𝜏) ∧ 𝜁) → 𝜎)
 
Theoremsyl3anr1 1408 A syllogism inference. (Contributed by NM, 31-Jul-2007.)
(𝜑𝜓)    &   ((𝜒 ∧ (𝜓𝜃𝜏)) → 𝜂)       ((𝜒 ∧ (𝜑𝜃𝜏)) → 𝜂)
 
Theoremsyl3anr2 1409 A syllogism inference. (Contributed by NM, 1-Aug-2007.) (Proof shortened by Wolf Lammen, 27-Jun-2022.)
(𝜑𝜃)    &   ((𝜒 ∧ (𝜓𝜃𝜏)) → 𝜂)       ((𝜒 ∧ (𝜓𝜑𝜏)) → 𝜂)
 
Theoremsyl3anr3 1410 A syllogism inference. (Contributed by NM, 23-Aug-2007.)
(𝜑𝜏)    &   ((𝜒 ∧ (𝜓𝜃𝜏)) → 𝜂)       ((𝜒 ∧ (𝜓𝜃𝜑)) → 𝜂)
 
Theorem3anidm12 1411 Inference from idempotent law for conjunction. (Contributed by NM, 7-Mar-2008.)
((𝜑𝜑𝜓) → 𝜒)       ((𝜑𝜓) → 𝜒)
 
Theorem3anidm13 1412 Inference from idempotent law for conjunction. (Contributed by NM, 7-Mar-2008.)
((𝜑𝜓𝜑) → 𝜒)       ((𝜑𝜓) → 𝜒)
 
Theorem3anidm23 1413 Inference from idempotent law for conjunction. (Contributed by NM, 1-Feb-2007.)
((𝜑𝜓𝜓) → 𝜒)       ((𝜑𝜓) → 𝜒)
 
Theoremsyl2an3an 1414 syl3an 1152 with antecedents in standard conjunction form. (Contributed by Alan Sare, 31-Aug-2016.)
(𝜑𝜓)    &   (𝜑𝜒)    &   (𝜃𝜏)    &   ((𝜓𝜒𝜏) → 𝜂)       ((𝜑𝜃) → 𝜂)
 
Theoremsyl2an23an 1415 Deduction related to syl3an 1152 with antecedents in standard conjunction form. (Contributed by Alan Sare, 31-Aug-2016.) (Proof shortened by Wolf Lammen, 28-Jun-2022.)
(𝜑𝜓)    &   (𝜑𝜒)    &   ((𝜃𝜑) → 𝜏)    &   ((𝜓𝜒𝜏) → 𝜂)       ((𝜃𝜑) → 𝜂)
 
Theorem3ori 1416 Infer implication from triple disjunction. (Contributed by NM, 26-Sep-2006.)
(𝜑𝜓𝜒)       ((¬ 𝜑 ∧ ¬ 𝜓) → 𝜒)
 
Theorem3jao 1417 Disjunction of three antecedents. (Contributed by NM, 8-Apr-1994.)
(((𝜑𝜓) ∧ (𝜒𝜓) ∧ (𝜃𝜓)) → ((𝜑𝜒𝜃) → 𝜓))
 
Theorem3jaob 1418 Disjunction of three antecedents. (Contributed by NM, 13-Sep-2011.)
(((𝜑𝜒𝜃) → 𝜓) ↔ ((𝜑𝜓) ∧ (𝜒𝜓) ∧ (𝜃𝜓)))
 
Theorem3jaoi 1419 Disjunction of three antecedents (inference). (Contributed by NM, 12-Sep-1995.)
(𝜑𝜓)    &   (𝜒𝜓)    &   (𝜃𝜓)       ((𝜑𝜒𝜃) → 𝜓)
 
Theorem3jaod 1420 Disjunction of three antecedents (deduction). (Contributed by NM, 14-Oct-2005.)
(𝜑 → (𝜓𝜒))    &   (𝜑 → (𝜃𝜒))    &   (𝜑 → (𝜏𝜒))       (𝜑 → ((𝜓𝜃𝜏) → 𝜒))
 
Theorem3jaoian 1421 Disjunction of three antecedents (inference). (Contributed by NM, 14-Oct-2005.)
((𝜑𝜓) → 𝜒)    &   ((𝜃𝜓) → 𝜒)    &   ((𝜏𝜓) → 𝜒)       (((𝜑𝜃𝜏) ∧ 𝜓) → 𝜒)
 
Theorem3jaodan 1422 Disjunction of three antecedents (deduction). (Contributed by NM, 14-Oct-2005.)
((𝜑𝜓) → 𝜒)    &   ((𝜑𝜃) → 𝜒)    &   ((𝜑𝜏) → 𝜒)       ((𝜑 ∧ (𝜓𝜃𝜏)) → 𝜒)
 
Theoremmpjao3dan 1423 Eliminate a three-way disjunction in a deduction. (Contributed by Thierry Arnoux, 13-Apr-2018.)
((𝜑𝜓) → 𝜒)    &   ((𝜑𝜃) → 𝜒)    &   ((𝜑𝜏) → 𝜒)    &   (𝜑 → (𝜓𝜃𝜏))       (𝜑𝜒)
 
Theorem3jaao 1424 Inference conjoining and disjoining the antecedents of three implications. (Contributed by Jeff Hankins, 15-Aug-2009.) (Proof shortened by Andrew Salmon, 13-May-2011.)
(𝜑 → (𝜓𝜒))    &   (𝜃 → (𝜏𝜒))    &   (𝜂 → (𝜁𝜒))       ((𝜑𝜃𝜂) → ((𝜓𝜏𝜁) → 𝜒))
 
Theoremsyl3an9b 1425 Nested syllogism inference conjoining 3 dissimilar antecedents. (Contributed by NM, 1-May-1995.)
(𝜑 → (𝜓𝜒))    &   (𝜃 → (𝜒𝜏))    &   (𝜂 → (𝜏𝜁))       ((𝜑𝜃𝜂) → (𝜓𝜁))
 
Theorem3orbi123d 1426 Deduction joining 3 equivalences to form equivalence of disjunctions. (Contributed by NM, 20-Apr-1994.)
(𝜑 → (𝜓𝜒))    &   (𝜑 → (𝜃𝜏))    &   (𝜑 → (𝜂𝜁))       (𝜑 → ((𝜓𝜃𝜂) ↔ (𝜒𝜏𝜁)))
 
Theorem3anbi123d 1427 Deduction joining 3 equivalences to form equivalence of conjunctions. (Contributed by NM, 22-Apr-1994.)
(𝜑 → (𝜓𝜒))    &   (𝜑 → (𝜃𝜏))    &   (𝜑 → (𝜂𝜁))       (𝜑 → ((𝜓𝜃𝜂) ↔ (𝜒𝜏𝜁)))
 
Theorem3anbi12d 1428 Deduction conjoining and adding a conjunct to equivalences. (Contributed by NM, 8-Sep-2006.)
(𝜑 → (𝜓𝜒))    &   (𝜑 → (𝜃𝜏))       (𝜑 → ((𝜓𝜃𝜂) ↔ (𝜒𝜏𝜂)))
 
Theorem3anbi13d 1429 Deduction conjoining and adding a conjunct to equivalences. (Contributed by NM, 8-Sep-2006.)
(𝜑 → (𝜓𝜒))    &   (𝜑 → (𝜃𝜏))       (𝜑 → ((𝜓𝜂𝜃) ↔ (𝜒𝜂𝜏)))
 
Theorem3anbi23d 1430 Deduction conjoining and adding a conjunct to equivalences. (Contributed by NM, 8-Sep-2006.)
(𝜑 → (𝜓𝜒))    &   (𝜑 → (𝜃𝜏))       (𝜑 → ((𝜂𝜓𝜃) ↔ (𝜂𝜒𝜏)))
 
Theorem3anbi1d 1431 Deduction adding conjuncts to an equivalence. (Contributed by NM, 8-Sep-2006.)
(𝜑 → (𝜓𝜒))       (𝜑 → ((𝜓𝜃𝜏) ↔ (𝜒𝜃𝜏)))
 
Theorem3anbi2d 1432 Deduction adding conjuncts to an equivalence. (Contributed by NM, 8-Sep-2006.)
(𝜑 → (𝜓𝜒))       (𝜑 → ((𝜃𝜓𝜏) ↔ (𝜃𝜒𝜏)))
 
Theorem3anbi3d 1433 Deduction adding conjuncts to an equivalence. (Contributed by NM, 8-Sep-2006.)
(𝜑 → (𝜓𝜒))       (𝜑 → ((𝜃𝜏𝜓) ↔ (𝜃𝜏𝜒)))
 
Theorem3anim123d 1434 Deduction joining 3 implications to form implication of conjunctions. (Contributed by NM, 24-Feb-2005.)
(𝜑 → (𝜓𝜒))    &   (𝜑 → (𝜃𝜏))    &   (𝜑 → (𝜂𝜁))       (𝜑 → ((𝜓𝜃𝜂) → (𝜒𝜏𝜁)))
 
Theorem3orim123d 1435 Deduction joining 3 implications to form implication of disjunctions. (Contributed by NM, 4-Apr-1997.)
(𝜑 → (𝜓𝜒))    &   (𝜑 → (𝜃𝜏))    &   (𝜑 → (𝜂𝜁))       (𝜑 → ((𝜓𝜃𝜂) → (𝜒𝜏𝜁)))
 
Theoreman6 1436 Rearrangement of 6 conjuncts. (Contributed by NM, 13-Mar-1995.)
(((𝜑𝜓𝜒) ∧ (𝜃𝜏𝜂)) ↔ ((𝜑𝜃) ∧ (𝜓𝜏) ∧ (𝜒𝜂)))
 
Theorem3an6 1437 Analogue of an4 652 for triple conjunction. (Contributed by Scott Fenton, 16-Mar-2011.) (Proof shortened by Andrew Salmon, 25-May-2011.)
(((𝜑𝜓) ∧ (𝜒𝜃) ∧ (𝜏𝜂)) ↔ ((𝜑𝜒𝜏) ∧ (𝜓𝜃𝜂)))
 
Theorem3or6 1438 Analogue of or4 920 for triple conjunction. (Contributed by Scott Fenton, 16-Mar-2011.)
(((𝜑𝜓) ∨ (𝜒𝜃) ∨ (𝜏𝜂)) ↔ ((𝜑𝜒𝜏) ∨ (𝜓𝜃𝜂)))
 
Theoremmp3an1 1439 An inference based on modus ponens. (Contributed by NM, 21-Nov-1994.)
𝜑    &   ((𝜑𝜓𝜒) → 𝜃)       ((𝜓𝜒) → 𝜃)
 
Theoremmp3an2 1440 An inference based on modus ponens. (Contributed by NM, 21-Nov-1994.)
𝜓    &   ((𝜑𝜓𝜒) → 𝜃)       ((𝜑𝜒) → 𝜃)
 
Theoremmp3an3 1441 An inference based on modus ponens. (Contributed by NM, 21-Nov-1994.)
𝜒    &   ((𝜑𝜓𝜒) → 𝜃)       ((𝜑𝜓) → 𝜃)
 
Theoremmp3an12 1442 An inference based on modus ponens. (Contributed by NM, 13-Jul-2005.)
𝜑    &   𝜓    &   ((𝜑𝜓𝜒) → 𝜃)       (𝜒𝜃)
 
Theoremmp3an13 1443 An inference based on modus ponens. (Contributed by NM, 14-Jul-2005.)
𝜑    &   𝜒    &   ((𝜑𝜓𝜒) → 𝜃)       (𝜓𝜃)
 
Theoremmp3an23 1444 An inference based on modus ponens. (Contributed by NM, 14-Jul-2005.)
𝜓    &   𝜒    &   ((𝜑𝜓𝜒) → 𝜃)       (𝜑𝜃)
 
Theoremmp3an1i 1445 An inference based on modus ponens. (Contributed by NM, 5-Jul-2005.)
𝜓    &   (𝜑 → ((𝜓𝜒𝜃) → 𝜏))       (𝜑 → ((𝜒𝜃) → 𝜏))
 
Theoremmp3anl1 1446 An inference based on modus ponens. (Contributed by NM, 24-Feb-2005.)
𝜑    &   (((𝜑𝜓𝜒) ∧ 𝜃) → 𝜏)       (((𝜓𝜒) ∧ 𝜃) → 𝜏)
 
Theoremmp3anl2 1447 An inference based on modus ponens. (Contributed by NM, 24-Feb-2005.)
𝜓    &   (((𝜑𝜓𝜒) ∧ 𝜃) → 𝜏)       (((𝜑𝜒) ∧ 𝜃) → 𝜏)
 
Theoremmp3anl3 1448 An inference based on modus ponens. (Contributed by NM, 24-Feb-2005.)
𝜒    &   (((𝜑𝜓𝜒) ∧ 𝜃) → 𝜏)       (((𝜑𝜓) ∧ 𝜃) → 𝜏)
 
Theoremmp3anr1 1449 An inference based on modus ponens. (Contributed by NM, 4-Nov-2006.)
𝜓    &   ((𝜑 ∧ (𝜓𝜒𝜃)) → 𝜏)       ((𝜑 ∧ (𝜒𝜃)) → 𝜏)
 
Theoremmp3anr2 1450 An inference based on modus ponens. (Contributed by NM, 24-Nov-2006.)
𝜒    &   ((𝜑 ∧ (𝜓𝜒𝜃)) → 𝜏)       ((𝜑 ∧ (𝜓𝜃)) → 𝜏)
 
Theoremmp3anr3 1451 An inference based on modus ponens. (Contributed by NM, 19-Oct-2007.)
𝜃    &   ((𝜑 ∧ (𝜓𝜒𝜃)) → 𝜏)       ((𝜑 ∧ (𝜓𝜒)) → 𝜏)
 
Theoremmp3an 1452 An inference based on modus ponens. (Contributed by NM, 14-May-1999.)
𝜑    &   𝜓    &   𝜒    &   ((𝜑𝜓𝜒) → 𝜃)       𝜃
 
Theoremmpd3an3 1453 An inference based on modus ponens. (Contributed by NM, 8-Nov-2007.)
((𝜑𝜓) → 𝜒)    &   ((𝜑𝜓𝜒) → 𝜃)       ((𝜑𝜓) → 𝜃)
 
Theoremmpd3an23 1454 An inference based on modus ponens. (Contributed by NM, 4-Dec-2006.)
(𝜑𝜓)    &   (𝜑𝜒)    &   ((𝜑𝜓𝜒) → 𝜃)       (𝜑𝜃)
 
Theoremmp3and 1455 A deduction based on modus ponens. (Contributed by Mario Carneiro, 24-Dec-2016.)
(𝜑𝜓)    &   (𝜑𝜒)    &   (𝜑𝜃)    &   (𝜑 → ((𝜓𝜒𝜃) → 𝜏))       (𝜑𝜏)
 
Theoremmp3an12i 1456 mp3an 1452 with antecedents in standard conjunction form and with one hypothesis an implication. (Contributed by Alan Sare, 28-Aug-2016.)
𝜑    &   𝜓    &   (𝜒𝜃)    &   ((𝜑𝜓𝜃) → 𝜏)       (𝜒𝜏)
 
Theoremmp3an2i 1457 mp3an 1452 with antecedents in standard conjunction form and with two hypotheses which are implications. (Contributed by Alan Sare, 28-Aug-2016.)
𝜑    &   (𝜓𝜒)    &   (𝜓𝜃)    &   ((𝜑𝜒𝜃) → 𝜏)       (𝜓𝜏)
 
Theoremmp3an3an 1458 mp3an 1452 with antecedents in standard conjunction form and with two hypotheses which are implications. (Contributed by Alan Sare, 28-Aug-2016.)
𝜑    &   (𝜓𝜒)    &   (𝜃𝜏)    &   ((𝜑𝜒𝜏) → 𝜂)       ((𝜓𝜃) → 𝜂)
 
Theoremmp3an2ani 1459 An elimination deduction. (Contributed by Alan Sare, 17-Oct-2017.)
𝜑    &   (𝜓𝜒)    &   ((𝜓𝜃) → 𝜏)    &   ((𝜑𝜒𝜏) → 𝜂)       ((𝜓𝜃) → 𝜂)
 
Theorembiimp3a 1460 Infer implication from a logical equivalence. Similar to biimpa 477. (Contributed by NM, 4-Sep-2005.)
((𝜑𝜓) → (𝜒𝜃))       ((𝜑𝜓𝜒) → 𝜃)
 
Theorembiimp3ar 1461 Infer implication from a logical equivalence. Similar to biimpar 478. (Contributed by NM, 2-Jan-2009.)
((𝜑𝜓) → (𝜒𝜃))       ((𝜑𝜓𝜃) → 𝜒)
 
Theorem3anandis 1462 Inference that undistributes a triple conjunction in the antecedent. (Contributed by NM, 18-Apr-2007.)
(((𝜑𝜓) ∧ (𝜑𝜒) ∧ (𝜑𝜃)) → 𝜏)       ((𝜑 ∧ (𝜓𝜒𝜃)) → 𝜏)
 
Theorem3anandirs 1463 Inference that undistributes a triple conjunction in the antecedent. (Contributed by NM, 25-Jul-2006.)
(((𝜑𝜃) ∧ (𝜓𝜃) ∧ (𝜒𝜃)) → 𝜏)       (((𝜑𝜓𝜒) ∧ 𝜃) → 𝜏)
 
Theoremecase23d 1464 Deduction for elimination by cases. (Contributed by NM, 22-Apr-1994.)
(𝜑 → ¬ 𝜒)    &   (𝜑 → ¬ 𝜃)    &   (𝜑 → (𝜓𝜒𝜃))       (𝜑𝜓)
 
Theorem3ecase 1465 Inference for elimination by cases. (Contributed by NM, 13-Jul-2005.)
𝜑𝜃)    &   𝜓𝜃)    &   𝜒𝜃)    &   ((𝜑𝜓𝜒) → 𝜃)       𝜃
 
Theorem3bior1fd 1466 A disjunction is equivalent to a threefold disjunction with single falsehood, analogous to biorf 930. (Contributed by Alexander van der Vekens, 8-Sep-2017.)
(𝜑 → ¬ 𝜃)       (𝜑 → ((𝜒𝜓) ↔ (𝜃𝜒𝜓)))
 
Theorem3bior1fand 1467 A disjunction is equivalent to a threefold disjunction with single falsehood of a conjunction. (Contributed by Alexander van der Vekens, 8-Sep-2017.)
(𝜑 → ¬ 𝜃)       (𝜑 → ((𝜒𝜓) ↔ ((𝜃𝜏) ∨ 𝜒𝜓)))
 
Theorem3bior2fd 1468 A wff is equivalent to its threefold disjunction with double falsehood, analogous to biorf 930. (Contributed by Alexander van der Vekens, 8-Sep-2017.)
(𝜑 → ¬ 𝜃)    &   (𝜑 → ¬ 𝜒)       (𝜑 → (𝜓 ↔ (𝜃𝜒𝜓)))
 
Theorem3biant1d 1469 A conjunction is equivalent to a threefold conjunction with single truth, analogous to biantrud 532. (Contributed by Alexander van der Vekens, 26-Sep-2017.)
(𝜑𝜃)       (𝜑 → ((𝜒𝜓) ↔ (𝜃𝜒𝜓)))
 
Theoremintn3an1d 1470 Introduction of a triple conjunct inside a contradiction. (Contributed by FL, 27-Dec-2007.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
(𝜑 → ¬ 𝜓)       (𝜑 → ¬ (𝜓𝜒𝜃))
 
Theoremintn3an2d 1471 Introduction of a triple conjunct inside a contradiction. (Contributed by FL, 27-Dec-2007.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
(𝜑 → ¬ 𝜓)       (𝜑 → ¬ (𝜒𝜓𝜃))
 
Theoremintn3an3d 1472 Introduction of a triple conjunct inside a contradiction. (Contributed by FL, 27-Dec-2007.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)
(𝜑 → ¬ 𝜓)       (𝜑 → ¬ (𝜒𝜃𝜓))
 
Theoreman3andi 1473 Distribution of conjunction over threefold conjunction. (Contributed by Thierry Arnoux, 8-Apr-2019.)
((𝜑 ∧ (𝜓𝜒𝜃)) ↔ ((𝜑𝜓) ∧ (𝜑𝜒) ∧ (𝜑𝜃)))
 
Theoreman33rean 1474 Rearrange a 9-fold conjunction. (Contributed by Thierry Arnoux, 14-Apr-2019.)
(((𝜑𝜓𝜒) ∧ (𝜃𝜏𝜂) ∧ (𝜁𝜎𝜌)) ↔ ((𝜑𝜏𝜌) ∧ ((𝜓𝜃) ∧ (𝜂𝜎) ∧ (𝜒𝜁))))
 
1.2.12  Logical "nand" (Sheffer stroke)
 
Syntaxwnan 1475 Extend wff definition to include alternative denial ("nand").
wff (𝜑𝜓)
 
Definitiondf-nan 1476 Define incompatibility, or alternative denial ("not-and" or "nand"). This is also called the Sheffer stroke, represented by a vertical bar, but we use a different symbol to avoid ambiguity with other uses of the vertical bar. In the second edition of Principia Mathematica (1927), Russell and Whitehead used the Sheffer stroke and suggested it as a replacement for the "or" and "not" operations of the first edition. However, in practice, "or" and "not" are more widely used. After we define the constant true (df-tru 1531) and the constant false (df-fal 1541), we will be able to prove these truth table values: ((⊤ ⊼ ⊤) ↔ ⊥) (trunantru 1569), ((⊤ ⊼ ⊥) ↔ ⊤) (trunanfal 1570), ((⊥ ⊼ ⊤) ↔ ⊤) (falnantru 1571), and ((⊥ ⊼ ⊥) ↔ ⊤) (falnanfal 1572). Contrast with (df-an 397), (df-or 842), (wi 4), and (df-xor 1496). (Contributed by Jeff Hoffman, 19-Nov-2007.)
((𝜑𝜓) ↔ ¬ (𝜑𝜓))
 
Theoremnanan 1477 Conjunction in terms of alternative denial. (Contributed by Mario Carneiro, 9-May-2015.)
((𝜑𝜓) ↔ ¬ (𝜑𝜓))
 
Theoremnanimn 1478 Alternative denial in terms of our primitive connectives (implication and negation). (Contributed by WL, 26-Jun-2020.)
((𝜑𝜓) ↔ (𝜑 → ¬ 𝜓))
 
Theoremnanor 1479 Alternative denial in terms of disjunction and negation. This explains the name "alternative denial". (Contributed by BJ, 19-Oct-2022.)
((𝜑𝜓) ↔ (¬ 𝜑 ∨ ¬ 𝜓))
 
Theoremnancom 1480 Alternative denial is commutative. Remark: alternative denial is not associative, see nanass 1494. (Contributed by Mario Carneiro, 9-May-2015.) (Proof shortened by Wolf Lammen, 26-Jun-2020.)
((𝜑𝜓) ↔ (𝜓𝜑))
 
Theoremnannan 1481 Nested alternative denials. (Contributed by Jeff Hoffman, 19-Nov-2007.) (Proof shortened by Wolf Lammen, 26-Jun-2020.)
((𝜑 ⊼ (𝜓𝜒)) ↔ (𝜑 → (𝜓𝜒)))
 
Theoremnanim 1482 Implication in terms of alternative denial. (Contributed by Jeff Hoffman, 19-Nov-2007.)
((𝜑𝜓) ↔ (𝜑 ⊼ (𝜓𝜓)))
 
Theoremnannot 1483 Negation in terms of alternative denial. (Contributed by Jeff Hoffman, 19-Nov-2007.) (Revised by Wolf Lammen, 26-Jun-2020.)
𝜑 ↔ (𝜑𝜑))
 
Theoremnanbi 1484 Biconditional in terms of alternative denial. (Contributed by Jeff Hoffman, 19-Nov-2007.) (Proof shortened by Wolf Lammen, 27-Jun-2020.)
((𝜑𝜓) ↔ ((𝜑𝜓) ⊼ ((𝜑𝜑) ⊼ (𝜓𝜓))))
 
Theoremnanbi1 1485 Introduce a right anti-conjunct to both sides of a logical equivalence. (Contributed by Anthony Hart, 1-Sep-2011.) (Proof shortened by Wolf Lammen, 27-Jun-2020.)
((𝜑𝜓) → ((𝜑𝜒) ↔ (𝜓𝜒)))
 
Theoremnanbi2 1486 Introduce a left anti-conjunct to both sides of a logical equivalence. (Contributed by Anthony Hart, 1-Sep-2011.) (Proof shortened by SF, 2-Jan-2018.)
((𝜑𝜓) → ((𝜒𝜑) ↔ (𝜒𝜓)))
 
Theoremnanbi12 1487 Join two logical equivalences with anti-conjunction. (Contributed by SF, 2-Jan-2018.)
(((𝜑𝜓) ∧ (𝜒𝜃)) → ((𝜑𝜒) ↔ (𝜓𝜃)))
 
Theoremnanbi1i 1488 Introduce a right anti-conjunct to both sides of a logical equivalence. (Contributed by SF, 2-Jan-2018.)
(𝜑𝜓)       ((𝜑𝜒) ↔ (𝜓𝜒))
 
Theoremnanbi2i 1489 Introduce a left anti-conjunct to both sides of a logical equivalence. (Contributed by SF, 2-Jan-2018.)
(𝜑𝜓)       ((𝜒𝜑) ↔ (𝜒𝜓))
 
Theoremnanbi12i 1490 Join two logical equivalences with anti-conjunction. (Contributed by SF, 2-Jan-2018.)
(𝜑𝜓)    &   (𝜒𝜃)       ((𝜑𝜒) ↔ (𝜓𝜃))
 
Theoremnanbi1d 1491 Introduce a right anti-conjunct to both sides of a logical equivalence. (Contributed by SF, 2-Jan-2018.)
(𝜑 → (𝜓𝜒))       (𝜑 → ((𝜓𝜃) ↔ (𝜒𝜃)))
 
Theoremnanbi2d 1492 Introduce a left anti-conjunct to both sides of a logical equivalence. (Contributed by SF, 2-Jan-2018.)
(𝜑 → (𝜓𝜒))       (𝜑 → ((𝜃𝜓) ↔ (𝜃𝜒)))
 
Theoremnanbi12d 1493 Join two logical equivalences with anti-conjunction. (Contributed by Scott Fenton, 2-Jan-2018.)
(𝜑 → (𝜓𝜒))    &   (𝜑 → (𝜃𝜏))       (𝜑 → ((𝜓𝜃) ↔ (𝜒𝜏)))
 
Theoremnanass 1494 A characterization of when an expression involving alternative denials associates. Remark: alternative denial is commutative, see nancom 1480. (Contributed by Richard Penner, 29-Feb-2020.) (Proof shortened by Wolf Lammen, 23-Oct-2022.)
((𝜑𝜒) ↔ (((𝜑𝜓) ⊼ 𝜒) ↔ (𝜑 ⊼ (𝜓𝜒))))
 
1.2.13  Logical "xor"
 
Syntaxwxo 1495 Extend wff definition to include exclusive disjunction ("xor").
wff (𝜑𝜓)
 
Definitiondf-xor 1496 Define exclusive disjunction (logical "xor"). Return true if either the left or right, but not both, are true. After we define the constant true (df-tru 1531) and the constant false (df-fal 1541), we will be able to prove these truth table values: ((⊤ ⊻ ⊤) ↔ ⊥) (truxortru 1573), ((⊤ ⊻ ⊥) ↔ ⊤) (truxorfal 1574), ((⊥ ⊻ ⊤) ↔ ⊤) (falxortru 1575), and ((⊥ ⊻ ⊥) ↔ ⊥) (falxorfal 1576). Contrast with (df-an 397), (df-or 842), (wi 4), and (df-nan 1476). (Contributed by FL, 22-Nov-2010.)
((𝜑𝜓) ↔ ¬ (𝜑𝜓))
 
Theoremxnor 1497 Two ways to write XNOR. (Contributed by Mario Carneiro, 4-Sep-2016.)
((𝜑𝜓) ↔ ¬ (𝜑𝜓))
 
Theoremxorcom 1498 The connector is commutative. (Contributed by Mario Carneiro, 4-Sep-2016.)
((𝜑𝜓) ↔ (𝜓𝜑))
 
Theoremxorass 1499 The connector is associative. (Contributed by FL, 22-Nov-2010.) (Proof shortened by Andrew Salmon, 8-Jun-2011.) (Proof shortened by Wolf Lammen, 20-Jun-2020.)
(((𝜑𝜓) ⊻ 𝜒) ↔ (𝜑 ⊻ (𝜓𝜒)))
 
Theoremexcxor 1500 This tautology shows that xor is really exclusive. (Contributed by FL, 22-Nov-2010.)
((𝜑𝜓) ↔ ((𝜑 ∧ ¬ 𝜓) ∨ (¬ 𝜑𝜓)))
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