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Theorem List for Metamath Proof Explorer - 1401-1500   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremsyl3an2b 1401 A syllogism inference. (Contributed by NM, 22-Aug-1995.)
(𝜑𝜒)    &   ((𝜓𝜒𝜃) → 𝜏)       ((𝜓𝜑𝜃) → 𝜏)

Theoremsyl3an3b 1402 A syllogism inference. (Contributed by NM, 22-Aug-1995.)
(𝜑𝜃)    &   ((𝜓𝜒𝜃) → 𝜏)       ((𝜓𝜒𝜑) → 𝜏)

Theoremsyl3an1br 1403 A syllogism inference. (Contributed by NM, 22-Aug-1995.)
(𝜓𝜑)    &   ((𝜓𝜒𝜃) → 𝜏)       ((𝜑𝜒𝜃) → 𝜏)

Theoremsyl3an2br 1404 A syllogism inference. (Contributed by NM, 22-Aug-1995.)
(𝜒𝜑)    &   ((𝜓𝜒𝜃) → 𝜏)       ((𝜓𝜑𝜃) → 𝜏)

Theoremsyl3an3br 1405 A syllogism inference. (Contributed by NM, 22-Aug-1995.)
(𝜃𝜑)    &   ((𝜓𝜒𝜃) → 𝜏)       ((𝜓𝜒𝜑) → 𝜏)

Theoremsyld3an3 1406 A syllogism inference. (Contributed by NM, 20-May-2007.)
((𝜑𝜓𝜒) → 𝜃)    &   ((𝜑𝜓𝜃) → 𝜏)       ((𝜑𝜓𝜒) → 𝜏)

Theoremsyld3an1 1407 A syllogism inference. (Contributed by NM, 7-Jul-2008.) (Proof shortened by Wolf Lammen, 26-Jun-2022.)
((𝜒𝜓𝜃) → 𝜑)    &   ((𝜑𝜓𝜃) → 𝜏)       ((𝜒𝜓𝜃) → 𝜏)

Theoremsyld3an2 1408 A syllogism inference. (Contributed by NM, 20-May-2007.)
((𝜑𝜒𝜃) → 𝜓)    &   ((𝜑𝜓𝜃) → 𝜏)       ((𝜑𝜒𝜃) → 𝜏)

Theoremsyl3anl1 1409 A syllogism inference. (Contributed by NM, 24-Feb-2005.)
(𝜑𝜓)    &   (((𝜓𝜒𝜃) ∧ 𝜏) → 𝜂)       (((𝜑𝜒𝜃) ∧ 𝜏) → 𝜂)

Theoremsyl3anl2 1410 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 1411 A syllogism inference. (Contributed by NM, 24-Feb-2005.)
(𝜑𝜃)    &   (((𝜓𝜒𝜃) ∧ 𝜏) → 𝜂)       (((𝜓𝜒𝜑) ∧ 𝜏) → 𝜂)

Theoremsyl3anl 1412 A triple syllogism inference. (Contributed by NM, 24-Dec-2006.)
(𝜑𝜓)    &   (𝜒𝜃)    &   (𝜏𝜂)    &   (((𝜓𝜃𝜂) ∧ 𝜁) → 𝜎)       (((𝜑𝜒𝜏) ∧ 𝜁) → 𝜎)

Theoremsyl3anr1 1413 A syllogism inference. (Contributed by NM, 31-Jul-2007.)
(𝜑𝜓)    &   ((𝜒 ∧ (𝜓𝜃𝜏)) → 𝜂)       ((𝜒 ∧ (𝜑𝜃𝜏)) → 𝜂)

Theoremsyl3anr2 1414 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 1415 A syllogism inference. (Contributed by NM, 23-Aug-2007.)
(𝜑𝜏)    &   ((𝜒 ∧ (𝜓𝜃𝜏)) → 𝜂)       ((𝜒 ∧ (𝜓𝜃𝜑)) → 𝜂)

Theorem3anidm12 1416 Inference from idempotent law for conjunction. (Contributed by NM, 7-Mar-2008.)
((𝜑𝜑𝜓) → 𝜒)       ((𝜑𝜓) → 𝜒)

Theorem3anidm13 1417 Inference from idempotent law for conjunction. (Contributed by NM, 7-Mar-2008.)
((𝜑𝜓𝜑) → 𝜒)       ((𝜑𝜓) → 𝜒)

Theorem3anidm23 1418 Inference from idempotent law for conjunction. (Contributed by NM, 1-Feb-2007.)
((𝜑𝜓𝜓) → 𝜒)       ((𝜑𝜓) → 𝜒)

Theoremsyl2an3an 1419 syl3an 1157 with antecedents in standard conjunction form. (Contributed by Alan Sare, 31-Aug-2016.)
(𝜑𝜓)    &   (𝜑𝜒)    &   (𝜃𝜏)    &   ((𝜓𝜒𝜏) → 𝜂)       ((𝜑𝜃) → 𝜂)

Theoremsyl2an23an 1420 Deduction related to syl3an 1157 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 1421 Infer implication from triple disjunction. (Contributed by NM, 26-Sep-2006.)
(𝜑𝜓𝜒)       ((¬ 𝜑 ∧ ¬ 𝜓) → 𝜒)

Theorem3jao 1422 Disjunction of three antecedents. (Contributed by NM, 8-Apr-1994.)
(((𝜑𝜓) ∧ (𝜒𝜓) ∧ (𝜃𝜓)) → ((𝜑𝜒𝜃) → 𝜓))

Theorem3jaob 1423 Disjunction of three antecedents. (Contributed by NM, 13-Sep-2011.)
(((𝜑𝜒𝜃) → 𝜓) ↔ ((𝜑𝜓) ∧ (𝜒𝜓) ∧ (𝜃𝜓)))

Theorem3jaoi 1424 Disjunction of three antecedents (inference). (Contributed by NM, 12-Sep-1995.)
(𝜑𝜓)    &   (𝜒𝜓)    &   (𝜃𝜓)       ((𝜑𝜒𝜃) → 𝜓)

Theorem3jaod 1425 Disjunction of three antecedents (deduction). (Contributed by NM, 14-Oct-2005.)
(𝜑 → (𝜓𝜒))    &   (𝜑 → (𝜃𝜒))    &   (𝜑 → (𝜏𝜒))       (𝜑 → ((𝜓𝜃𝜏) → 𝜒))

Theorem3jaoian 1426 Disjunction of three antecedents (inference). (Contributed by NM, 14-Oct-2005.)
((𝜑𝜓) → 𝜒)    &   ((𝜃𝜓) → 𝜒)    &   ((𝜏𝜓) → 𝜒)       (((𝜑𝜃𝜏) ∧ 𝜓) → 𝜒)

Theorem3jaodan 1427 Disjunction of three antecedents (deduction). (Contributed by NM, 14-Oct-2005.)
((𝜑𝜓) → 𝜒)    &   ((𝜑𝜃) → 𝜒)    &   ((𝜑𝜏) → 𝜒)       ((𝜑 ∧ (𝜓𝜃𝜏)) → 𝜒)

Theoremmpjao3dan 1428 Eliminate a three-way disjunction in a deduction. (Contributed by Thierry Arnoux, 13-Apr-2018.) (Proof shortened by Wolf Lammen, 20-Apr-2024.)
((𝜑𝜓) → 𝜒)    &   ((𝜑𝜃) → 𝜒)    &   ((𝜑𝜏) → 𝜒)    &   (𝜑 → (𝜓𝜃𝜏))       (𝜑𝜒)

Theoremmpjao3danOLD 1429 Obsolete version of mpjao3dan 1428 as of 17-Apr-2024. (Contributed by Thierry Arnoux, 13-Apr-2018.) (New usage is discouraged.) (Proof modification is discouraged.)
((𝜑𝜓) → 𝜒)    &   ((𝜑𝜃) → 𝜒)    &   ((𝜑𝜏) → 𝜒)    &   (𝜑 → (𝜓𝜃𝜏))       (𝜑𝜒)

Theorem3jaao 1430 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 1431 Nested syllogism inference conjoining 3 dissimilar antecedents. (Contributed by NM, 1-May-1995.)
(𝜑 → (𝜓𝜒))    &   (𝜃 → (𝜒𝜏))    &   (𝜂 → (𝜏𝜁))       ((𝜑𝜃𝜂) → (𝜓𝜁))

Theorem3orbi123d 1432 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 1433 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 1434 Deduction conjoining and adding a conjunct to equivalences. (Contributed by NM, 8-Sep-2006.)
(𝜑 → (𝜓𝜒))    &   (𝜑 → (𝜃𝜏))       (𝜑 → ((𝜓𝜃𝜂) ↔ (𝜒𝜏𝜂)))

Theorem3anbi13d 1435 Deduction conjoining and adding a conjunct to equivalences. (Contributed by NM, 8-Sep-2006.)
(𝜑 → (𝜓𝜒))    &   (𝜑 → (𝜃𝜏))       (𝜑 → ((𝜓𝜂𝜃) ↔ (𝜒𝜂𝜏)))

Theorem3anbi23d 1436 Deduction conjoining and adding a conjunct to equivalences. (Contributed by NM, 8-Sep-2006.)
(𝜑 → (𝜓𝜒))    &   (𝜑 → (𝜃𝜏))       (𝜑 → ((𝜂𝜓𝜃) ↔ (𝜂𝜒𝜏)))

Theorem3anbi1d 1437 Deduction adding conjuncts to an equivalence. (Contributed by NM, 8-Sep-2006.)
(𝜑 → (𝜓𝜒))       (𝜑 → ((𝜓𝜃𝜏) ↔ (𝜒𝜃𝜏)))

Theorem3anbi2d 1438 Deduction adding conjuncts to an equivalence. (Contributed by NM, 8-Sep-2006.)
(𝜑 → (𝜓𝜒))       (𝜑 → ((𝜃𝜓𝜏) ↔ (𝜃𝜒𝜏)))

Theorem3anbi3d 1439 Deduction adding conjuncts to an equivalence. (Contributed by NM, 8-Sep-2006.)
(𝜑 → (𝜓𝜒))       (𝜑 → ((𝜃𝜏𝜓) ↔ (𝜃𝜏𝜒)))

Theorem3anim123d 1440 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 1441 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 1442 Rearrangement of 6 conjuncts. (Contributed by NM, 13-Mar-1995.)
(((𝜑𝜓𝜒) ∧ (𝜃𝜏𝜂)) ↔ ((𝜑𝜃) ∧ (𝜓𝜏) ∧ (𝜒𝜂)))

Theorem3an6 1443 Analogue of an4 655 for triple conjunction. (Contributed by Scott Fenton, 16-Mar-2011.) (Proof shortened by Andrew Salmon, 25-May-2011.)
(((𝜑𝜓) ∧ (𝜒𝜃) ∧ (𝜏𝜂)) ↔ ((𝜑𝜒𝜏) ∧ (𝜓𝜃𝜂)))

Theorem3or6 1444 Analogue of or4 924 for triple conjunction. (Contributed by Scott Fenton, 16-Mar-2011.)
(((𝜑𝜓) ∨ (𝜒𝜃) ∨ (𝜏𝜂)) ↔ ((𝜑𝜒𝜏) ∨ (𝜓𝜃𝜂)))

Theoremmp3an1 1445 An inference based on modus ponens. (Contributed by NM, 21-Nov-1994.)
𝜑    &   ((𝜑𝜓𝜒) → 𝜃)       ((𝜓𝜒) → 𝜃)

Theoremmp3an2 1446 An inference based on modus ponens. (Contributed by NM, 21-Nov-1994.)
𝜓    &   ((𝜑𝜓𝜒) → 𝜃)       ((𝜑𝜒) → 𝜃)

Theoremmp3an3 1447 An inference based on modus ponens. (Contributed by NM, 21-Nov-1994.)
𝜒    &   ((𝜑𝜓𝜒) → 𝜃)       ((𝜑𝜓) → 𝜃)

Theoremmp3an12 1448 An inference based on modus ponens. (Contributed by NM, 13-Jul-2005.)
𝜑    &   𝜓    &   ((𝜑𝜓𝜒) → 𝜃)       (𝜒𝜃)

Theoremmp3an13 1449 An inference based on modus ponens. (Contributed by NM, 14-Jul-2005.)
𝜑    &   𝜒    &   ((𝜑𝜓𝜒) → 𝜃)       (𝜓𝜃)

Theoremmp3an23 1450 An inference based on modus ponens. (Contributed by NM, 14-Jul-2005.)
𝜓    &   𝜒    &   ((𝜑𝜓𝜒) → 𝜃)       (𝜑𝜃)

Theoremmp3an1i 1451 An inference based on modus ponens. (Contributed by NM, 5-Jul-2005.)
𝜓    &   (𝜑 → ((𝜓𝜒𝜃) → 𝜏))       (𝜑 → ((𝜒𝜃) → 𝜏))

Theoremmp3anl1 1452 An inference based on modus ponens. (Contributed by NM, 24-Feb-2005.)
𝜑    &   (((𝜑𝜓𝜒) ∧ 𝜃) → 𝜏)       (((𝜓𝜒) ∧ 𝜃) → 𝜏)

Theoremmp3anl2 1453 An inference based on modus ponens. (Contributed by NM, 24-Feb-2005.)
𝜓    &   (((𝜑𝜓𝜒) ∧ 𝜃) → 𝜏)       (((𝜑𝜒) ∧ 𝜃) → 𝜏)

Theoremmp3anl3 1454 An inference based on modus ponens. (Contributed by NM, 24-Feb-2005.)
𝜒    &   (((𝜑𝜓𝜒) ∧ 𝜃) → 𝜏)       (((𝜑𝜓) ∧ 𝜃) → 𝜏)

Theoremmp3anr1 1455 An inference based on modus ponens. (Contributed by NM, 4-Nov-2006.)
𝜓    &   ((𝜑 ∧ (𝜓𝜒𝜃)) → 𝜏)       ((𝜑 ∧ (𝜒𝜃)) → 𝜏)

Theoremmp3anr2 1456 An inference based on modus ponens. (Contributed by NM, 24-Nov-2006.)
𝜒    &   ((𝜑 ∧ (𝜓𝜒𝜃)) → 𝜏)       ((𝜑 ∧ (𝜓𝜃)) → 𝜏)

Theoremmp3anr3 1457 An inference based on modus ponens. (Contributed by NM, 19-Oct-2007.)
𝜃    &   ((𝜑 ∧ (𝜓𝜒𝜃)) → 𝜏)       ((𝜑 ∧ (𝜓𝜒)) → 𝜏)

Theoremmp3an 1458 An inference based on modus ponens. (Contributed by NM, 14-May-1999.)
𝜑    &   𝜓    &   𝜒    &   ((𝜑𝜓𝜒) → 𝜃)       𝜃

Theoremmpd3an3 1459 An inference based on modus ponens. (Contributed by NM, 8-Nov-2007.)
((𝜑𝜓) → 𝜒)    &   ((𝜑𝜓𝜒) → 𝜃)       ((𝜑𝜓) → 𝜃)

Theoremmpd3an23 1460 An inference based on modus ponens. (Contributed by NM, 4-Dec-2006.)
(𝜑𝜓)    &   (𝜑𝜒)    &   ((𝜑𝜓𝜒) → 𝜃)       (𝜑𝜃)

Theoremmp3and 1461 A deduction based on modus ponens. (Contributed by Mario Carneiro, 24-Dec-2016.)
(𝜑𝜓)    &   (𝜑𝜒)    &   (𝜑𝜃)    &   (𝜑 → ((𝜓𝜒𝜃) → 𝜏))       (𝜑𝜏)

Theoremmp3an12i 1462 mp3an 1458 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 1463 mp3an 1458 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 1464 mp3an 1458 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 1465 An elimination deduction. (Contributed by Alan Sare, 17-Oct-2017.)
𝜑    &   (𝜓𝜒)    &   ((𝜓𝜃) → 𝜏)    &   ((𝜑𝜒𝜏) → 𝜂)       ((𝜓𝜃) → 𝜂)

Theorembiimp3a 1466 Infer implication from a logical equivalence. Similar to biimpa 480. (Contributed by NM, 4-Sep-2005.)
((𝜑𝜓) → (𝜒𝜃))       ((𝜑𝜓𝜒) → 𝜃)

Theorembiimp3ar 1467 Infer implication from a logical equivalence. Similar to biimpar 481. (Contributed by NM, 2-Jan-2009.)
((𝜑𝜓) → (𝜒𝜃))       ((𝜑𝜓𝜃) → 𝜒)

Theorem3anandis 1468 Inference that undistributes a triple conjunction in the antecedent. (Contributed by NM, 18-Apr-2007.)
(((𝜑𝜓) ∧ (𝜑𝜒) ∧ (𝜑𝜃)) → 𝜏)       ((𝜑 ∧ (𝜓𝜒𝜃)) → 𝜏)

Theorem3anandirs 1469 Inference that undistributes a triple conjunction in the antecedent. (Contributed by NM, 25-Jul-2006.)
(((𝜑𝜃) ∧ (𝜓𝜃) ∧ (𝜒𝜃)) → 𝜏)       (((𝜑𝜓𝜒) ∧ 𝜃) → 𝜏)

Theoremecase23d 1470 Deduction for elimination by cases. (Contributed by NM, 22-Apr-1994.)
(𝜑 → ¬ 𝜒)    &   (𝜑 → ¬ 𝜃)    &   (𝜑 → (𝜓𝜒𝜃))       (𝜑𝜓)

Theorem3ecase 1471 Inference for elimination by cases. (Contributed by NM, 13-Jul-2005.)
𝜑𝜃)    &   𝜓𝜃)    &   𝜒𝜃)    &   ((𝜑𝜓𝜒) → 𝜃)       𝜃

Theorem3bior1fd 1472 A disjunction is equivalent to a threefold disjunction with single falsehood, analogous to biorf 934. (Contributed by Alexander van der Vekens, 8-Sep-2017.)
(𝜑 → ¬ 𝜃)       (𝜑 → ((𝜒𝜓) ↔ (𝜃𝜒𝜓)))

Theorem3bior1fand 1473 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 1474 A wff is equivalent to its threefold disjunction with double falsehood, analogous to biorf 934. (Contributed by Alexander van der Vekens, 8-Sep-2017.)
(𝜑 → ¬ 𝜃)    &   (𝜑 → ¬ 𝜒)       (𝜑 → (𝜓 ↔ (𝜃𝜒𝜓)))

Theorem3biant1d 1475 A conjunction is equivalent to a threefold conjunction with single truth, analogous to biantrud 535. (Contributed by Alexander van der Vekens, 26-Sep-2017.)
(𝜑𝜃)       (𝜑 → ((𝜒𝜓) ↔ (𝜃𝜒𝜓)))

Theoremintn3an1d 1476 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 1477 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 1478 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 1479 Distribution of conjunction over threefold conjunction. (Contributed by Thierry Arnoux, 8-Apr-2019.)
((𝜑 ∧ (𝜓𝜒𝜃)) ↔ ((𝜑𝜓) ∧ (𝜑𝜒) ∧ (𝜑𝜃)))

Theoreman33rean 1480 Rearrange a 9-fold conjunction. (Contributed by Thierry Arnoux, 14-Apr-2019.) (Proof shortened by Wolf Lammen, 21-Apr-2024.)
(((𝜑𝜓𝜒) ∧ (𝜃𝜏𝜂) ∧ (𝜁𝜎𝜌)) ↔ ((𝜑𝜏𝜌) ∧ ((𝜓𝜃) ∧ (𝜂𝜎) ∧ (𝜒𝜁))))

Theoreman33reanOLD 1481 Obsolete version of an33rean 1480 as of 21-Apr-2024. (Contributed by Thierry Arnoux, 14-Apr-2019.) (Proof modification is discouraged.) (New usage is discouraged.)
(((𝜑𝜓𝜒) ∧ (𝜃𝜏𝜂) ∧ (𝜁𝜎𝜌)) ↔ ((𝜑𝜏𝜌) ∧ ((𝜓𝜃) ∧ (𝜂𝜎) ∧ (𝜒𝜁))))

1.2.12  Logical "nand" (Sheffer stroke)

Syntaxwnan 1482 Extend wff definition to include alternative denial ("nand").
wff (𝜑𝜓)

Definitiondf-nan 1483 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 1541) and the constant false (df-fal 1551), we will be able to prove these truth table values: ((⊤ ⊼ ⊤) ↔ ⊥) (trunantru 1579), ((⊤ ⊼ ⊥) ↔ ⊤) (trunanfal 1580), ((⊥ ⊼ ⊤) ↔ ⊤) (falnantru 1581), and ((⊥ ⊼ ⊥) ↔ ⊤) (falnanfal 1582). Contrast with (df-an 400), (df-or 845), (wi 4), and (df-xor 1503). (Contributed by Jeff Hoffman, 19-Nov-2007.)
((𝜑𝜓) ↔ ¬ (𝜑𝜓))

Theoremnanan 1484 Conjunction in terms of alternative denial. (Contributed by Mario Carneiro, 9-May-2015.)
((𝜑𝜓) ↔ ¬ (𝜑𝜓))

Theoremnanimn 1485 Alternative denial in terms of our primitive connectives (implication and negation). (Contributed by WL, 26-Jun-2020.)
((𝜑𝜓) ↔ (𝜑 → ¬ 𝜓))

Theoremnanor 1486 Alternative denial in terms of disjunction and negation. This explains the name "alternative denial". (Contributed by BJ, 19-Oct-2022.)
((𝜑𝜓) ↔ (¬ 𝜑 ∨ ¬ 𝜓))

Theoremnancom 1487 Alternative denial is commutative. Remark: alternative denial is not associative, see nanass 1501. (Contributed by Mario Carneiro, 9-May-2015.) (Proof shortened by Wolf Lammen, 26-Jun-2020.)
((𝜑𝜓) ↔ (𝜓𝜑))

Theoremnannan 1488 Nested alternative denials. (Contributed by Jeff Hoffman, 19-Nov-2007.) (Proof shortened by Wolf Lammen, 26-Jun-2020.)
((𝜑 ⊼ (𝜓𝜒)) ↔ (𝜑 → (𝜓𝜒)))

Theoremnanim 1489 Implication in terms of alternative denial. (Contributed by Jeff Hoffman, 19-Nov-2007.)
((𝜑𝜓) ↔ (𝜑 ⊼ (𝜓𝜓)))

Theoremnannot 1490 Negation in terms of alternative denial. (Contributed by Jeff Hoffman, 19-Nov-2007.) (Revised by Wolf Lammen, 26-Jun-2020.)
𝜑 ↔ (𝜑𝜑))

Theoremnanbi 1491 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 1492 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 1493 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 1494 Join two logical equivalences with anti-conjunction. (Contributed by SF, 2-Jan-2018.)
(((𝜑𝜓) ∧ (𝜒𝜃)) → ((𝜑𝜒) ↔ (𝜓𝜃)))

Theoremnanbi1i 1495 Introduce a right anti-conjunct to both sides of a logical equivalence. (Contributed by SF, 2-Jan-2018.)
(𝜑𝜓)       ((𝜑𝜒) ↔ (𝜓𝜒))

Theoremnanbi2i 1496 Introduce a left anti-conjunct to both sides of a logical equivalence. (Contributed by SF, 2-Jan-2018.)
(𝜑𝜓)       ((𝜒𝜑) ↔ (𝜒𝜓))

Theoremnanbi12i 1497 Join two logical equivalences with anti-conjunction. (Contributed by SF, 2-Jan-2018.)
(𝜑𝜓)    &   (𝜒𝜃)       ((𝜑𝜒) ↔ (𝜓𝜃))

Theoremnanbi1d 1498 Introduce a right anti-conjunct to both sides of a logical equivalence. (Contributed by SF, 2-Jan-2018.)
(𝜑 → (𝜓𝜒))       (𝜑 → ((𝜓𝜃) ↔ (𝜒𝜃)))

Theoremnanbi2d 1499 Introduce a left anti-conjunct to both sides of a logical equivalence. (Contributed by SF, 2-Jan-2018.)
(𝜑 → (𝜓𝜒))       (𝜑 → ((𝜃𝜓) ↔ (𝜃𝜒)))

Theoremnanbi12d 1500 Join two logical equivalences with anti-conjunction. (Contributed by Scott Fenton, 2-Jan-2018.)
(𝜑 → (𝜓𝜒))    &   (𝜑 → (𝜃𝜏))       (𝜑 → ((𝜓𝜃) ↔ (𝜒𝜏)))

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