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Type | Label | Description |
---|---|---|
Statement | ||
Theorem | syld3an3 1401 | A syllogism inference. (Contributed by NM, 20-May-2007.) |
⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜃) → 𝜏) ⇒ ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜏) | ||
Theorem | syld3an1 1402 | A syllogism inference. (Contributed by NM, 7-Jul-2008.) (Proof shortened by Wolf Lammen, 26-Jun-2022.) |
⊢ ((𝜒 ∧ 𝜓 ∧ 𝜃) → 𝜑) & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜃) → 𝜏) ⇒ ⊢ ((𝜒 ∧ 𝜓 ∧ 𝜃) → 𝜏) | ||
Theorem | syld3an2 1403 | A syllogism inference. (Contributed by NM, 20-May-2007.) |
⊢ ((𝜑 ∧ 𝜒 ∧ 𝜃) → 𝜓) & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜃) → 𝜏) ⇒ ⊢ ((𝜑 ∧ 𝜒 ∧ 𝜃) → 𝜏) | ||
Theorem | syl3anl1 1404 | A syllogism inference. (Contributed by NM, 24-Feb-2005.) |
⊢ (𝜑 → 𝜓) & ⊢ (((𝜓 ∧ 𝜒 ∧ 𝜃) ∧ 𝜏) → 𝜂) ⇒ ⊢ (((𝜑 ∧ 𝜒 ∧ 𝜃) ∧ 𝜏) → 𝜂) | ||
Theorem | syl3anl2 1405 | A syllogism inference. (Contributed by NM, 24-Feb-2005.) (Proof shortened by Wolf Lammen, 27-Jun-2022.) |
⊢ (𝜑 → 𝜒) & ⊢ (((𝜓 ∧ 𝜒 ∧ 𝜃) ∧ 𝜏) → 𝜂) ⇒ ⊢ (((𝜓 ∧ 𝜑 ∧ 𝜃) ∧ 𝜏) → 𝜂) | ||
Theorem | syl3anl3 1406 | A syllogism inference. (Contributed by NM, 24-Feb-2005.) |
⊢ (𝜑 → 𝜃) & ⊢ (((𝜓 ∧ 𝜒 ∧ 𝜃) ∧ 𝜏) → 𝜂) ⇒ ⊢ (((𝜓 ∧ 𝜒 ∧ 𝜑) ∧ 𝜏) → 𝜂) | ||
Theorem | syl3anl 1407 | A triple syllogism inference. (Contributed by NM, 24-Dec-2006.) |
⊢ (𝜑 → 𝜓) & ⊢ (𝜒 → 𝜃) & ⊢ (𝜏 → 𝜂) & ⊢ (((𝜓 ∧ 𝜃 ∧ 𝜂) ∧ 𝜁) → 𝜎) ⇒ ⊢ (((𝜑 ∧ 𝜒 ∧ 𝜏) ∧ 𝜁) → 𝜎) | ||
Theorem | syl3anr1 1408 | A syllogism inference. (Contributed by NM, 31-Jul-2007.) |
⊢ (𝜑 → 𝜓) & ⊢ ((𝜒 ∧ (𝜓 ∧ 𝜃 ∧ 𝜏)) → 𝜂) ⇒ ⊢ ((𝜒 ∧ (𝜑 ∧ 𝜃 ∧ 𝜏)) → 𝜂) | ||
Theorem | syl3anr2 1409 | A syllogism inference. (Contributed by NM, 1-Aug-2007.) (Proof shortened by Wolf Lammen, 27-Jun-2022.) |
⊢ (𝜑 → 𝜃) & ⊢ ((𝜒 ∧ (𝜓 ∧ 𝜃 ∧ 𝜏)) → 𝜂) ⇒ ⊢ ((𝜒 ∧ (𝜓 ∧ 𝜑 ∧ 𝜏)) → 𝜂) | ||
Theorem | syl3anr3 1410 | A syllogism inference. (Contributed by NM, 23-Aug-2007.) |
⊢ (𝜑 → 𝜏) & ⊢ ((𝜒 ∧ (𝜓 ∧ 𝜃 ∧ 𝜏)) → 𝜂) ⇒ ⊢ ((𝜒 ∧ (𝜓 ∧ 𝜃 ∧ 𝜑)) → 𝜂) | ||
Theorem | 3anidm12 1411 | Inference from idempotent law for conjunction. (Contributed by NM, 7-Mar-2008.) |
⊢ ((𝜑 ∧ 𝜑 ∧ 𝜓) → 𝜒) ⇒ ⊢ ((𝜑 ∧ 𝜓) → 𝜒) | ||
Theorem | 3anidm13 1412 | Inference from idempotent law for conjunction. (Contributed by NM, 7-Mar-2008.) |
⊢ ((𝜑 ∧ 𝜓 ∧ 𝜑) → 𝜒) ⇒ ⊢ ((𝜑 ∧ 𝜓) → 𝜒) | ||
Theorem | 3anidm23 1413 | Inference from idempotent law for conjunction. (Contributed by NM, 1-Feb-2007.) |
⊢ ((𝜑 ∧ 𝜓 ∧ 𝜓) → 𝜒) ⇒ ⊢ ((𝜑 ∧ 𝜓) → 𝜒) | ||
Theorem | syl2an3an 1414 | syl3an 1152 with antecedents in standard conjunction form. (Contributed by Alan Sare, 31-Aug-2016.) |
⊢ (𝜑 → 𝜓) & ⊢ (𝜑 → 𝜒) & ⊢ (𝜃 → 𝜏) & ⊢ ((𝜓 ∧ 𝜒 ∧ 𝜏) → 𝜂) ⇒ ⊢ ((𝜑 ∧ 𝜃) → 𝜂) | ||
Theorem | syl2an23an 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.) |
⊢ (𝜑 → 𝜓) & ⊢ (𝜑 → 𝜒) & ⊢ ((𝜃 ∧ 𝜑) → 𝜏) & ⊢ ((𝜓 ∧ 𝜒 ∧ 𝜏) → 𝜂) ⇒ ⊢ ((𝜃 ∧ 𝜑) → 𝜂) | ||
Theorem | 3ori 1416 | Infer implication from triple disjunction. (Contributed by NM, 26-Sep-2006.) |
⊢ (𝜑 ∨ 𝜓 ∨ 𝜒) ⇒ ⊢ ((¬ 𝜑 ∧ ¬ 𝜓) → 𝜒) | ||
Theorem | 3jao 1417 | Disjunction of three antecedents. (Contributed by NM, 8-Apr-1994.) |
⊢ (((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)) → ((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓)) | ||
Theorem | 3jaob 1418 | Disjunction of three antecedents. (Contributed by NM, 13-Sep-2011.) |
⊢ (((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓) ↔ ((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓))) | ||
Theorem | 3jaoi 1419 | Disjunction of three antecedents (inference). (Contributed by NM, 12-Sep-1995.) |
⊢ (𝜑 → 𝜓) & ⊢ (𝜒 → 𝜓) & ⊢ (𝜃 → 𝜓) ⇒ ⊢ ((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓) | ||
Theorem | 3jaod 1420 | Disjunction of three antecedents (deduction). (Contributed by NM, 14-Oct-2005.) |
⊢ (𝜑 → (𝜓 → 𝜒)) & ⊢ (𝜑 → (𝜃 → 𝜒)) & ⊢ (𝜑 → (𝜏 → 𝜒)) ⇒ ⊢ (𝜑 → ((𝜓 ∨ 𝜃 ∨ 𝜏) → 𝜒)) | ||
Theorem | 3jaoian 1421 | Disjunction of three antecedents (inference). (Contributed by NM, 14-Oct-2005.) |
⊢ ((𝜑 ∧ 𝜓) → 𝜒) & ⊢ ((𝜃 ∧ 𝜓) → 𝜒) & ⊢ ((𝜏 ∧ 𝜓) → 𝜒) ⇒ ⊢ (((𝜑 ∨ 𝜃 ∨ 𝜏) ∧ 𝜓) → 𝜒) | ||
Theorem | 3jaodan 1422 | Disjunction of three antecedents (deduction). (Contributed by NM, 14-Oct-2005.) |
⊢ ((𝜑 ∧ 𝜓) → 𝜒) & ⊢ ((𝜑 ∧ 𝜃) → 𝜒) & ⊢ ((𝜑 ∧ 𝜏) → 𝜒) ⇒ ⊢ ((𝜑 ∧ (𝜓 ∨ 𝜃 ∨ 𝜏)) → 𝜒) | ||
Theorem | mpjao3dan 1423 | Eliminate a three-way disjunction in a deduction. (Contributed by Thierry Arnoux, 13-Apr-2018.) |
⊢ ((𝜑 ∧ 𝜓) → 𝜒) & ⊢ ((𝜑 ∧ 𝜃) → 𝜒) & ⊢ ((𝜑 ∧ 𝜏) → 𝜒) & ⊢ (𝜑 → (𝜓 ∨ 𝜃 ∨ 𝜏)) ⇒ ⊢ (𝜑 → 𝜒) | ||
Theorem | 3jaao 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 → (𝜓 → 𝜒)) & ⊢ (𝜃 → (𝜏 → 𝜒)) & ⊢ (𝜂 → (𝜁 → 𝜒)) ⇒ ⊢ ((𝜑 ∧ 𝜃 ∧ 𝜂) → ((𝜓 ∨ 𝜏 ∨ 𝜁) → 𝜒)) | ||
Theorem | syl3an9b 1425 | Nested syllogism inference conjoining 3 dissimilar antecedents. (Contributed by NM, 1-May-1995.) |
⊢ (𝜑 → (𝜓 ↔ 𝜒)) & ⊢ (𝜃 → (𝜒 ↔ 𝜏)) & ⊢ (𝜂 → (𝜏 ↔ 𝜁)) ⇒ ⊢ ((𝜑 ∧ 𝜃 ∧ 𝜂) → (𝜓 ↔ 𝜁)) | ||
Theorem | 3orbi123d 1426 | Deduction joining 3 equivalences to form equivalence of disjunctions. (Contributed by NM, 20-Apr-1994.) |
⊢ (𝜑 → (𝜓 ↔ 𝜒)) & ⊢ (𝜑 → (𝜃 ↔ 𝜏)) & ⊢ (𝜑 → (𝜂 ↔ 𝜁)) ⇒ ⊢ (𝜑 → ((𝜓 ∨ 𝜃 ∨ 𝜂) ↔ (𝜒 ∨ 𝜏 ∨ 𝜁))) | ||
Theorem | 3anbi123d 1427 | Deduction joining 3 equivalences to form equivalence of conjunctions. (Contributed by NM, 22-Apr-1994.) |
⊢ (𝜑 → (𝜓 ↔ 𝜒)) & ⊢ (𝜑 → (𝜃 ↔ 𝜏)) & ⊢ (𝜑 → (𝜂 ↔ 𝜁)) ⇒ ⊢ (𝜑 → ((𝜓 ∧ 𝜃 ∧ 𝜂) ↔ (𝜒 ∧ 𝜏 ∧ 𝜁))) | ||
Theorem | 3anbi12d 1428 | Deduction conjoining and adding a conjunct to equivalences. (Contributed by NM, 8-Sep-2006.) |
⊢ (𝜑 → (𝜓 ↔ 𝜒)) & ⊢ (𝜑 → (𝜃 ↔ 𝜏)) ⇒ ⊢ (𝜑 → ((𝜓 ∧ 𝜃 ∧ 𝜂) ↔ (𝜒 ∧ 𝜏 ∧ 𝜂))) | ||
Theorem | 3anbi13d 1429 | Deduction conjoining and adding a conjunct to equivalences. (Contributed by NM, 8-Sep-2006.) |
⊢ (𝜑 → (𝜓 ↔ 𝜒)) & ⊢ (𝜑 → (𝜃 ↔ 𝜏)) ⇒ ⊢ (𝜑 → ((𝜓 ∧ 𝜂 ∧ 𝜃) ↔ (𝜒 ∧ 𝜂 ∧ 𝜏))) | ||
Theorem | 3anbi23d 1430 | Deduction conjoining and adding a conjunct to equivalences. (Contributed by NM, 8-Sep-2006.) |
⊢ (𝜑 → (𝜓 ↔ 𝜒)) & ⊢ (𝜑 → (𝜃 ↔ 𝜏)) ⇒ ⊢ (𝜑 → ((𝜂 ∧ 𝜓 ∧ 𝜃) ↔ (𝜂 ∧ 𝜒 ∧ 𝜏))) | ||
Theorem | 3anbi1d 1431 | Deduction adding conjuncts to an equivalence. (Contributed by NM, 8-Sep-2006.) |
⊢ (𝜑 → (𝜓 ↔ 𝜒)) ⇒ ⊢ (𝜑 → ((𝜓 ∧ 𝜃 ∧ 𝜏) ↔ (𝜒 ∧ 𝜃 ∧ 𝜏))) | ||
Theorem | 3anbi2d 1432 | Deduction adding conjuncts to an equivalence. (Contributed by NM, 8-Sep-2006.) |
⊢ (𝜑 → (𝜓 ↔ 𝜒)) ⇒ ⊢ (𝜑 → ((𝜃 ∧ 𝜓 ∧ 𝜏) ↔ (𝜃 ∧ 𝜒 ∧ 𝜏))) | ||
Theorem | 3anbi3d 1433 | Deduction adding conjuncts to an equivalence. (Contributed by NM, 8-Sep-2006.) |
⊢ (𝜑 → (𝜓 ↔ 𝜒)) ⇒ ⊢ (𝜑 → ((𝜃 ∧ 𝜏 ∧ 𝜓) ↔ (𝜃 ∧ 𝜏 ∧ 𝜒))) | ||
Theorem | 3anim123d 1434 | Deduction joining 3 implications to form implication of conjunctions. (Contributed by NM, 24-Feb-2005.) |
⊢ (𝜑 → (𝜓 → 𝜒)) & ⊢ (𝜑 → (𝜃 → 𝜏)) & ⊢ (𝜑 → (𝜂 → 𝜁)) ⇒ ⊢ (𝜑 → ((𝜓 ∧ 𝜃 ∧ 𝜂) → (𝜒 ∧ 𝜏 ∧ 𝜁))) | ||
Theorem | 3orim123d 1435 | Deduction joining 3 implications to form implication of disjunctions. (Contributed by NM, 4-Apr-1997.) |
⊢ (𝜑 → (𝜓 → 𝜒)) & ⊢ (𝜑 → (𝜃 → 𝜏)) & ⊢ (𝜑 → (𝜂 → 𝜁)) ⇒ ⊢ (𝜑 → ((𝜓 ∨ 𝜃 ∨ 𝜂) → (𝜒 ∨ 𝜏 ∨ 𝜁))) | ||
Theorem | an6 1436 | Rearrangement of 6 conjuncts. (Contributed by NM, 13-Mar-1995.) |
⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ∧ (𝜃 ∧ 𝜏 ∧ 𝜂)) ↔ ((𝜑 ∧ 𝜃) ∧ (𝜓 ∧ 𝜏) ∧ (𝜒 ∧ 𝜂))) | ||
Theorem | 3an6 1437 | Analogue of an4 652 for triple conjunction. (Contributed by Scott Fenton, 16-Mar-2011.) (Proof shortened by Andrew Salmon, 25-May-2011.) |
⊢ (((𝜑 ∧ 𝜓) ∧ (𝜒 ∧ 𝜃) ∧ (𝜏 ∧ 𝜂)) ↔ ((𝜑 ∧ 𝜒 ∧ 𝜏) ∧ (𝜓 ∧ 𝜃 ∧ 𝜂))) | ||
Theorem | 3or6 1438 | Analogue of or4 920 for triple conjunction. (Contributed by Scott Fenton, 16-Mar-2011.) |
⊢ (((𝜑 ∨ 𝜓) ∨ (𝜒 ∨ 𝜃) ∨ (𝜏 ∨ 𝜂)) ↔ ((𝜑 ∨ 𝜒 ∨ 𝜏) ∨ (𝜓 ∨ 𝜃 ∨ 𝜂))) | ||
Theorem | mp3an1 1439 | An inference based on modus ponens. (Contributed by NM, 21-Nov-1994.) |
⊢ 𝜑 & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ⇒ ⊢ ((𝜓 ∧ 𝜒) → 𝜃) | ||
Theorem | mp3an2 1440 | An inference based on modus ponens. (Contributed by NM, 21-Nov-1994.) |
⊢ 𝜓 & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ⇒ ⊢ ((𝜑 ∧ 𝜒) → 𝜃) | ||
Theorem | mp3an3 1441 | An inference based on modus ponens. (Contributed by NM, 21-Nov-1994.) |
⊢ 𝜒 & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ⇒ ⊢ ((𝜑 ∧ 𝜓) → 𝜃) | ||
Theorem | mp3an12 1442 | An inference based on modus ponens. (Contributed by NM, 13-Jul-2005.) |
⊢ 𝜑 & ⊢ 𝜓 & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ⇒ ⊢ (𝜒 → 𝜃) | ||
Theorem | mp3an13 1443 | An inference based on modus ponens. (Contributed by NM, 14-Jul-2005.) |
⊢ 𝜑 & ⊢ 𝜒 & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ⇒ ⊢ (𝜓 → 𝜃) | ||
Theorem | mp3an23 1444 | An inference based on modus ponens. (Contributed by NM, 14-Jul-2005.) |
⊢ 𝜓 & ⊢ 𝜒 & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ⇒ ⊢ (𝜑 → 𝜃) | ||
Theorem | mp3an1i 1445 | An inference based on modus ponens. (Contributed by NM, 5-Jul-2005.) |
⊢ 𝜓 & ⊢ (𝜑 → ((𝜓 ∧ 𝜒 ∧ 𝜃) → 𝜏)) ⇒ ⊢ (𝜑 → ((𝜒 ∧ 𝜃) → 𝜏)) | ||
Theorem | mp3anl1 1446 | An inference based on modus ponens. (Contributed by NM, 24-Feb-2005.) |
⊢ 𝜑 & ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ∧ 𝜃) → 𝜏) ⇒ ⊢ (((𝜓 ∧ 𝜒) ∧ 𝜃) → 𝜏) | ||
Theorem | mp3anl2 1447 | An inference based on modus ponens. (Contributed by NM, 24-Feb-2005.) |
⊢ 𝜓 & ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ∧ 𝜃) → 𝜏) ⇒ ⊢ (((𝜑 ∧ 𝜒) ∧ 𝜃) → 𝜏) | ||
Theorem | mp3anl3 1448 | An inference based on modus ponens. (Contributed by NM, 24-Feb-2005.) |
⊢ 𝜒 & ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ∧ 𝜃) → 𝜏) ⇒ ⊢ (((𝜑 ∧ 𝜓) ∧ 𝜃) → 𝜏) | ||
Theorem | mp3anr1 1449 | An inference based on modus ponens. (Contributed by NM, 4-Nov-2006.) |
⊢ 𝜓 & ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒 ∧ 𝜃)) → 𝜏) ⇒ ⊢ ((𝜑 ∧ (𝜒 ∧ 𝜃)) → 𝜏) | ||
Theorem | mp3anr2 1450 | An inference based on modus ponens. (Contributed by NM, 24-Nov-2006.) |
⊢ 𝜒 & ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒 ∧ 𝜃)) → 𝜏) ⇒ ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜃)) → 𝜏) | ||
Theorem | mp3anr3 1451 | An inference based on modus ponens. (Contributed by NM, 19-Oct-2007.) |
⊢ 𝜃 & ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒 ∧ 𝜃)) → 𝜏) ⇒ ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒)) → 𝜏) | ||
Theorem | mp3an 1452 | An inference based on modus ponens. (Contributed by NM, 14-May-1999.) |
⊢ 𝜑 & ⊢ 𝜓 & ⊢ 𝜒 & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ⇒ ⊢ 𝜃 | ||
Theorem | mpd3an3 1453 | An inference based on modus ponens. (Contributed by NM, 8-Nov-2007.) |
⊢ ((𝜑 ∧ 𝜓) → 𝜒) & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ⇒ ⊢ ((𝜑 ∧ 𝜓) → 𝜃) | ||
Theorem | mpd3an23 1454 | An inference based on modus ponens. (Contributed by NM, 4-Dec-2006.) |
⊢ (𝜑 → 𝜓) & ⊢ (𝜑 → 𝜒) & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ⇒ ⊢ (𝜑 → 𝜃) | ||
Theorem | mp3and 1455 | A deduction based on modus ponens. (Contributed by Mario Carneiro, 24-Dec-2016.) |
⊢ (𝜑 → 𝜓) & ⊢ (𝜑 → 𝜒) & ⊢ (𝜑 → 𝜃) & ⊢ (𝜑 → ((𝜓 ∧ 𝜒 ∧ 𝜃) → 𝜏)) ⇒ ⊢ (𝜑 → 𝜏) | ||
Theorem | mp3an12i 1456 | mp3an 1452 with antecedents in standard conjunction form and with one hypothesis an implication. (Contributed by Alan Sare, 28-Aug-2016.) |
⊢ 𝜑 & ⊢ 𝜓 & ⊢ (𝜒 → 𝜃) & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜃) → 𝜏) ⇒ ⊢ (𝜒 → 𝜏) | ||
Theorem | mp3an2i 1457 | mp3an 1452 with antecedents in standard conjunction form and with two hypotheses which are implications. (Contributed by Alan Sare, 28-Aug-2016.) |
⊢ 𝜑 & ⊢ (𝜓 → 𝜒) & ⊢ (𝜓 → 𝜃) & ⊢ ((𝜑 ∧ 𝜒 ∧ 𝜃) → 𝜏) ⇒ ⊢ (𝜓 → 𝜏) | ||
Theorem | mp3an3an 1458 | mp3an 1452 with antecedents in standard conjunction form and with two hypotheses which are implications. (Contributed by Alan Sare, 28-Aug-2016.) |
⊢ 𝜑 & ⊢ (𝜓 → 𝜒) & ⊢ (𝜃 → 𝜏) & ⊢ ((𝜑 ∧ 𝜒 ∧ 𝜏) → 𝜂) ⇒ ⊢ ((𝜓 ∧ 𝜃) → 𝜂) | ||
Theorem | mp3an2ani 1459 | An elimination deduction. (Contributed by Alan Sare, 17-Oct-2017.) |
⊢ 𝜑 & ⊢ (𝜓 → 𝜒) & ⊢ ((𝜓 ∧ 𝜃) → 𝜏) & ⊢ ((𝜑 ∧ 𝜒 ∧ 𝜏) → 𝜂) ⇒ ⊢ ((𝜓 ∧ 𝜃) → 𝜂) | ||
Theorem | biimp3a 1460 | Infer implication from a logical equivalence. Similar to biimpa 477. (Contributed by NM, 4-Sep-2005.) |
⊢ ((𝜑 ∧ 𝜓) → (𝜒 ↔ 𝜃)) ⇒ ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) | ||
Theorem | biimp3ar 1461 | Infer implication from a logical equivalence. Similar to biimpar 478. (Contributed by NM, 2-Jan-2009.) |
⊢ ((𝜑 ∧ 𝜓) → (𝜒 ↔ 𝜃)) ⇒ ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜃) → 𝜒) | ||
Theorem | 3anandis 1462 | Inference that undistributes a triple conjunction in the antecedent. (Contributed by NM, 18-Apr-2007.) |
⊢ (((𝜑 ∧ 𝜓) ∧ (𝜑 ∧ 𝜒) ∧ (𝜑 ∧ 𝜃)) → 𝜏) ⇒ ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒 ∧ 𝜃)) → 𝜏) | ||
Theorem | 3anandirs 1463 | Inference that undistributes a triple conjunction in the antecedent. (Contributed by NM, 25-Jul-2006.) |
⊢ (((𝜑 ∧ 𝜃) ∧ (𝜓 ∧ 𝜃) ∧ (𝜒 ∧ 𝜃)) → 𝜏) ⇒ ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ∧ 𝜃) → 𝜏) | ||
Theorem | ecase23d 1464 | Deduction for elimination by cases. (Contributed by NM, 22-Apr-1994.) |
⊢ (𝜑 → ¬ 𝜒) & ⊢ (𝜑 → ¬ 𝜃) & ⊢ (𝜑 → (𝜓 ∨ 𝜒 ∨ 𝜃)) ⇒ ⊢ (𝜑 → 𝜓) | ||
Theorem | 3ecase 1465 | Inference for elimination by cases. (Contributed by NM, 13-Jul-2005.) |
⊢ (¬ 𝜑 → 𝜃) & ⊢ (¬ 𝜓 → 𝜃) & ⊢ (¬ 𝜒 → 𝜃) & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ⇒ ⊢ 𝜃 | ||
Theorem | 3bior1fd 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 → ((𝜒 ∨ 𝜓) ↔ (𝜃 ∨ 𝜒 ∨ 𝜓))) | ||
Theorem | 3bior1fand 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 → ((𝜒 ∨ 𝜓) ↔ ((𝜃 ∧ 𝜏) ∨ 𝜒 ∨ 𝜓))) | ||
Theorem | 3bior2fd 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 ↔ (𝜃 ∨ 𝜒 ∨ 𝜓))) | ||
Theorem | 3biant1d 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.) |
⊢ (𝜑 → 𝜃) ⇒ ⊢ (𝜑 → ((𝜒 ∧ 𝜓) ↔ (𝜃 ∧ 𝜒 ∧ 𝜓))) | ||
Theorem | intn3an1d 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 → ¬ (𝜓 ∧ 𝜒 ∧ 𝜃)) | ||
Theorem | intn3an2d 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 → ¬ (𝜒 ∧ 𝜓 ∧ 𝜃)) | ||
Theorem | intn3an3d 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 → ¬ (𝜒 ∧ 𝜃 ∧ 𝜓)) | ||
Theorem | an3andi 1473 | Distribution of conjunction over threefold conjunction. (Contributed by Thierry Arnoux, 8-Apr-2019.) |
⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒 ∧ 𝜃)) ↔ ((𝜑 ∧ 𝜓) ∧ (𝜑 ∧ 𝜒) ∧ (𝜑 ∧ 𝜃))) | ||
Theorem | an33rean 1474 | Rearrange a 9-fold conjunction. (Contributed by Thierry Arnoux, 14-Apr-2019.) |
⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ∧ (𝜃 ∧ 𝜏 ∧ 𝜂) ∧ (𝜁 ∧ 𝜎 ∧ 𝜌)) ↔ ((𝜑 ∧ 𝜏 ∧ 𝜌) ∧ ((𝜓 ∧ 𝜃) ∧ (𝜂 ∧ 𝜎) ∧ (𝜒 ∧ 𝜁)))) | ||
Syntax | wnan 1475 | Extend wff definition to include alternative denial ("nand"). |
wff (𝜑 ⊼ 𝜓) | ||
Definition | df-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.) |
⊢ ((𝜑 ⊼ 𝜓) ↔ ¬ (𝜑 ∧ 𝜓)) | ||
Theorem | nanan 1477 | Conjunction in terms of alternative denial. (Contributed by Mario Carneiro, 9-May-2015.) |
⊢ ((𝜑 ∧ 𝜓) ↔ ¬ (𝜑 ⊼ 𝜓)) | ||
Theorem | nanimn 1478 | Alternative denial in terms of our primitive connectives (implication and negation). (Contributed by WL, 26-Jun-2020.) |
⊢ ((𝜑 ⊼ 𝜓) ↔ (𝜑 → ¬ 𝜓)) | ||
Theorem | nanor 1479 | Alternative denial in terms of disjunction and negation. This explains the name "alternative denial". (Contributed by BJ, 19-Oct-2022.) |
⊢ ((𝜑 ⊼ 𝜓) ↔ (¬ 𝜑 ∨ ¬ 𝜓)) | ||
Theorem | nancom 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.) |
⊢ ((𝜑 ⊼ 𝜓) ↔ (𝜓 ⊼ 𝜑)) | ||
Theorem | nannan 1481 | Nested alternative denials. (Contributed by Jeff Hoffman, 19-Nov-2007.) (Proof shortened by Wolf Lammen, 26-Jun-2020.) |
⊢ ((𝜑 ⊼ (𝜓 ⊼ 𝜒)) ↔ (𝜑 → (𝜓 ∧ 𝜒))) | ||
Theorem | nanim 1482 | Implication in terms of alternative denial. (Contributed by Jeff Hoffman, 19-Nov-2007.) |
⊢ ((𝜑 → 𝜓) ↔ (𝜑 ⊼ (𝜓 ⊼ 𝜓))) | ||
Theorem | nannot 1483 | Negation in terms of alternative denial. (Contributed by Jeff Hoffman, 19-Nov-2007.) (Revised by Wolf Lammen, 26-Jun-2020.) |
⊢ (¬ 𝜑 ↔ (𝜑 ⊼ 𝜑)) | ||
Theorem | nanbi 1484 | Biconditional in terms of alternative denial. (Contributed by Jeff Hoffman, 19-Nov-2007.) (Proof shortened by Wolf Lammen, 27-Jun-2020.) |
⊢ ((𝜑 ↔ 𝜓) ↔ ((𝜑 ⊼ 𝜓) ⊼ ((𝜑 ⊼ 𝜑) ⊼ (𝜓 ⊼ 𝜓)))) | ||
Theorem | nanbi1 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.) |
⊢ ((𝜑 ↔ 𝜓) → ((𝜑 ⊼ 𝜒) ↔ (𝜓 ⊼ 𝜒))) | ||
Theorem | nanbi2 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.) |
⊢ ((𝜑 ↔ 𝜓) → ((𝜒 ⊼ 𝜑) ↔ (𝜒 ⊼ 𝜓))) | ||
Theorem | nanbi12 1487 | Join two logical equivalences with anti-conjunction. (Contributed by SF, 2-Jan-2018.) |
⊢ (((𝜑 ↔ 𝜓) ∧ (𝜒 ↔ 𝜃)) → ((𝜑 ⊼ 𝜒) ↔ (𝜓 ⊼ 𝜃))) | ||
Theorem | nanbi1i 1488 | Introduce a right anti-conjunct to both sides of a logical equivalence. (Contributed by SF, 2-Jan-2018.) |
⊢ (𝜑 ↔ 𝜓) ⇒ ⊢ ((𝜑 ⊼ 𝜒) ↔ (𝜓 ⊼ 𝜒)) | ||
Theorem | nanbi2i 1489 | Introduce a left anti-conjunct to both sides of a logical equivalence. (Contributed by SF, 2-Jan-2018.) |
⊢ (𝜑 ↔ 𝜓) ⇒ ⊢ ((𝜒 ⊼ 𝜑) ↔ (𝜒 ⊼ 𝜓)) | ||
Theorem | nanbi12i 1490 | Join two logical equivalences with anti-conjunction. (Contributed by SF, 2-Jan-2018.) |
⊢ (𝜑 ↔ 𝜓) & ⊢ (𝜒 ↔ 𝜃) ⇒ ⊢ ((𝜑 ⊼ 𝜒) ↔ (𝜓 ⊼ 𝜃)) | ||
Theorem | nanbi1d 1491 | Introduce a right anti-conjunct to both sides of a logical equivalence. (Contributed by SF, 2-Jan-2018.) |
⊢ (𝜑 → (𝜓 ↔ 𝜒)) ⇒ ⊢ (𝜑 → ((𝜓 ⊼ 𝜃) ↔ (𝜒 ⊼ 𝜃))) | ||
Theorem | nanbi2d 1492 | Introduce a left anti-conjunct to both sides of a logical equivalence. (Contributed by SF, 2-Jan-2018.) |
⊢ (𝜑 → (𝜓 ↔ 𝜒)) ⇒ ⊢ (𝜑 → ((𝜃 ⊼ 𝜓) ↔ (𝜃 ⊼ 𝜒))) | ||
Theorem | nanbi12d 1493 | Join two logical equivalences with anti-conjunction. (Contributed by Scott Fenton, 2-Jan-2018.) |
⊢ (𝜑 → (𝜓 ↔ 𝜒)) & ⊢ (𝜑 → (𝜃 ↔ 𝜏)) ⇒ ⊢ (𝜑 → ((𝜓 ⊼ 𝜃) ↔ (𝜒 ⊼ 𝜏))) | ||
Theorem | nanass 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.) |
⊢ ((𝜑 ↔ 𝜒) ↔ (((𝜑 ⊼ 𝜓) ⊼ 𝜒) ↔ (𝜑 ⊼ (𝜓 ⊼ 𝜒)))) | ||
Syntax | wxo 1495 | Extend wff definition to include exclusive disjunction ("xor"). |
wff (𝜑 ⊻ 𝜓) | ||
Definition | df-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.) |
⊢ ((𝜑 ⊻ 𝜓) ↔ ¬ (𝜑 ↔ 𝜓)) | ||
Theorem | xnor 1497 | Two ways to write XNOR. (Contributed by Mario Carneiro, 4-Sep-2016.) |
⊢ ((𝜑 ↔ 𝜓) ↔ ¬ (𝜑 ⊻ 𝜓)) | ||
Theorem | xorcom 1498 | The connector ⊻ is commutative. (Contributed by Mario Carneiro, 4-Sep-2016.) |
⊢ ((𝜑 ⊻ 𝜓) ↔ (𝜓 ⊻ 𝜑)) | ||
Theorem | xorass 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.) |
⊢ (((𝜑 ⊻ 𝜓) ⊻ 𝜒) ↔ (𝜑 ⊻ (𝜓 ⊻ 𝜒))) | ||
Theorem | excxor 1500 | This tautology shows that xor is really exclusive. (Contributed by FL, 22-Nov-2010.) |
⊢ ((𝜑 ⊻ 𝜓) ↔ ((𝜑 ∧ ¬ 𝜓) ∨ (¬ 𝜑 ∧ 𝜓))) |
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