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Type | Label | Description |
---|---|---|
Statement | ||
Theorem | syl323anc 1401 | Syllogism combined with contraction. (Contributed by NM, 11-Mar-2012.) |
⊢ (𝜑 → 𝜓) & ⊢ (𝜑 → 𝜒) & ⊢ (𝜑 → 𝜃) & ⊢ (𝜑 → 𝜏) & ⊢ (𝜑 → 𝜂) & ⊢ (𝜑 → 𝜁) & ⊢ (𝜑 → 𝜎) & ⊢ (𝜑 → 𝜌) & ⊢ (((𝜓 ∧ 𝜒 ∧ 𝜃) ∧ (𝜏 ∧ 𝜂) ∧ (𝜁 ∧ 𝜎 ∧ 𝜌)) → 𝜇) ⇒ ⊢ (𝜑 → 𝜇) | ||
Theorem | syl332anc 1402 | Syllogism combined with contraction. (Contributed by NM, 11-Mar-2012.) |
⊢ (𝜑 → 𝜓) & ⊢ (𝜑 → 𝜒) & ⊢ (𝜑 → 𝜃) & ⊢ (𝜑 → 𝜏) & ⊢ (𝜑 → 𝜂) & ⊢ (𝜑 → 𝜁) & ⊢ (𝜑 → 𝜎) & ⊢ (𝜑 → 𝜌) & ⊢ (((𝜓 ∧ 𝜒 ∧ 𝜃) ∧ (𝜏 ∧ 𝜂 ∧ 𝜁) ∧ (𝜎 ∧ 𝜌)) → 𝜇) ⇒ ⊢ (𝜑 → 𝜇) | ||
Theorem | syl333anc 1403 | A syllogism inference combined with contraction. (Contributed by NM, 10-Mar-2012.) |
⊢ (𝜑 → 𝜓) & ⊢ (𝜑 → 𝜒) & ⊢ (𝜑 → 𝜃) & ⊢ (𝜑 → 𝜏) & ⊢ (𝜑 → 𝜂) & ⊢ (𝜑 → 𝜁) & ⊢ (𝜑 → 𝜎) & ⊢ (𝜑 → 𝜌) & ⊢ (𝜑 → 𝜇) & ⊢ (((𝜓 ∧ 𝜒 ∧ 𝜃) ∧ (𝜏 ∧ 𝜂 ∧ 𝜁) ∧ (𝜎 ∧ 𝜌 ∧ 𝜇)) → 𝜆) ⇒ ⊢ (𝜑 → 𝜆) | ||
Theorem | syl3an1b 1404 | A syllogism inference. (Contributed by NM, 22-Aug-1995.) |
⊢ (𝜑 ↔ 𝜓) & ⊢ ((𝜓 ∧ 𝜒 ∧ 𝜃) → 𝜏) ⇒ ⊢ ((𝜑 ∧ 𝜒 ∧ 𝜃) → 𝜏) | ||
Theorem | syl3an2b 1405 | A syllogism inference. (Contributed by NM, 22-Aug-1995.) |
⊢ (𝜑 ↔ 𝜒) & ⊢ ((𝜓 ∧ 𝜒 ∧ 𝜃) → 𝜏) ⇒ ⊢ ((𝜓 ∧ 𝜑 ∧ 𝜃) → 𝜏) | ||
Theorem | syl3an3b 1406 | A syllogism inference. (Contributed by NM, 22-Aug-1995.) |
⊢ (𝜑 ↔ 𝜃) & ⊢ ((𝜓 ∧ 𝜒 ∧ 𝜃) → 𝜏) ⇒ ⊢ ((𝜓 ∧ 𝜒 ∧ 𝜑) → 𝜏) | ||
Theorem | syl3an1br 1407 | A syllogism inference. (Contributed by NM, 22-Aug-1995.) |
⊢ (𝜓 ↔ 𝜑) & ⊢ ((𝜓 ∧ 𝜒 ∧ 𝜃) → 𝜏) ⇒ ⊢ ((𝜑 ∧ 𝜒 ∧ 𝜃) → 𝜏) | ||
Theorem | syl3an2br 1408 | A syllogism inference. (Contributed by NM, 22-Aug-1995.) |
⊢ (𝜒 ↔ 𝜑) & ⊢ ((𝜓 ∧ 𝜒 ∧ 𝜃) → 𝜏) ⇒ ⊢ ((𝜓 ∧ 𝜑 ∧ 𝜃) → 𝜏) | ||
Theorem | syl3an3br 1409 | A syllogism inference. (Contributed by NM, 22-Aug-1995.) |
⊢ (𝜃 ↔ 𝜑) & ⊢ ((𝜓 ∧ 𝜒 ∧ 𝜃) → 𝜏) ⇒ ⊢ ((𝜓 ∧ 𝜒 ∧ 𝜑) → 𝜏) | ||
Theorem | syld3an3 1410 | A syllogism inference. (Contributed by NM, 20-May-2007.) |
⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜃) → 𝜏) ⇒ ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜏) | ||
Theorem | syld3an1 1411 | 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 1412 | A syllogism inference. (Contributed by NM, 20-May-2007.) |
⊢ ((𝜑 ∧ 𝜒 ∧ 𝜃) → 𝜓) & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜃) → 𝜏) ⇒ ⊢ ((𝜑 ∧ 𝜒 ∧ 𝜃) → 𝜏) | ||
Theorem | syl3anl1 1413 | A syllogism inference. (Contributed by NM, 24-Feb-2005.) |
⊢ (𝜑 → 𝜓) & ⊢ (((𝜓 ∧ 𝜒 ∧ 𝜃) ∧ 𝜏) → 𝜂) ⇒ ⊢ (((𝜑 ∧ 𝜒 ∧ 𝜃) ∧ 𝜏) → 𝜂) | ||
Theorem | syl3anl2 1414 | 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 1415 | A syllogism inference. (Contributed by NM, 24-Feb-2005.) |
⊢ (𝜑 → 𝜃) & ⊢ (((𝜓 ∧ 𝜒 ∧ 𝜃) ∧ 𝜏) → 𝜂) ⇒ ⊢ (((𝜓 ∧ 𝜒 ∧ 𝜑) ∧ 𝜏) → 𝜂) | ||
Theorem | syl3anl 1416 | 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 1417 | A syllogism inference. (Contributed by NM, 31-Jul-2007.) |
⊢ (𝜑 → 𝜓) & ⊢ ((𝜒 ∧ (𝜓 ∧ 𝜃 ∧ 𝜏)) → 𝜂) ⇒ ⊢ ((𝜒 ∧ (𝜑 ∧ 𝜃 ∧ 𝜏)) → 𝜂) | ||
Theorem | syl3anr2 1418 | 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 1419 | A syllogism inference. (Contributed by NM, 23-Aug-2007.) |
⊢ (𝜑 → 𝜏) & ⊢ ((𝜒 ∧ (𝜓 ∧ 𝜃 ∧ 𝜏)) → 𝜂) ⇒ ⊢ ((𝜒 ∧ (𝜓 ∧ 𝜃 ∧ 𝜑)) → 𝜂) | ||
Theorem | 3anidm12 1420 | Inference from idempotent law for conjunction. (Contributed by NM, 7-Mar-2008.) |
⊢ ((𝜑 ∧ 𝜑 ∧ 𝜓) → 𝜒) ⇒ ⊢ ((𝜑 ∧ 𝜓) → 𝜒) | ||
Theorem | 3anidm13 1421 | Inference from idempotent law for conjunction. (Contributed by NM, 7-Mar-2008.) |
⊢ ((𝜑 ∧ 𝜓 ∧ 𝜑) → 𝜒) ⇒ ⊢ ((𝜑 ∧ 𝜓) → 𝜒) | ||
Theorem | 3anidm23 1422 | Inference from idempotent law for conjunction. (Contributed by NM, 1-Feb-2007.) |
⊢ ((𝜑 ∧ 𝜓 ∧ 𝜓) → 𝜒) ⇒ ⊢ ((𝜑 ∧ 𝜓) → 𝜒) | ||
Theorem | syl2an3an 1423 | syl3an 1161 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 1424 | Deduction related to syl3an 1161 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 1425 | Infer implication from triple disjunction. (Contributed by NM, 26-Sep-2006.) |
⊢ (𝜑 ∨ 𝜓 ∨ 𝜒) ⇒ ⊢ ((¬ 𝜑 ∧ ¬ 𝜓) → 𝜒) | ||
Theorem | 3jao 1426 | Disjunction of three antecedents. (Contributed by NM, 8-Apr-1994.) |
⊢ (((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)) → ((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓)) | ||
Theorem | 3jaob 1427 | Disjunction of three antecedents. (Contributed by NM, 13-Sep-2011.) |
⊢ (((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓) ↔ ((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓))) | ||
Theorem | 3jaoi 1428 | Disjunction of three antecedents (inference). (Contributed by NM, 12-Sep-1995.) |
⊢ (𝜑 → 𝜓) & ⊢ (𝜒 → 𝜓) & ⊢ (𝜃 → 𝜓) ⇒ ⊢ ((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓) | ||
Theorem | 3jaod 1429 | 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 1430 | 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 1431 | 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 1432 | Eliminate a three-way disjunction in a deduction. (Contributed by Thierry Arnoux, 13-Apr-2018.) (Proof shortened by Wolf Lammen, 20-Apr-2024.) |
⊢ ((𝜑 ∧ 𝜓) → 𝜒) & ⊢ ((𝜑 ∧ 𝜃) → 𝜒) & ⊢ ((𝜑 ∧ 𝜏) → 𝜒) & ⊢ (𝜑 → (𝜓 ∨ 𝜃 ∨ 𝜏)) ⇒ ⊢ (𝜑 → 𝜒) | ||
Theorem | mpjao3danOLD 1433 | Obsolete version of mpjao3dan 1432 as of 17-Apr-2024. (Contributed by Thierry Arnoux, 13-Apr-2018.) (New usage is discouraged.) (Proof modification is discouraged.) |
⊢ ((𝜑 ∧ 𝜓) → 𝜒) & ⊢ ((𝜑 ∧ 𝜃) → 𝜒) & ⊢ ((𝜑 ∧ 𝜏) → 𝜒) & ⊢ (𝜑 → (𝜓 ∨ 𝜃 ∨ 𝜏)) ⇒ ⊢ (𝜑 → 𝜒) | ||
Theorem | 3jaao 1434 | 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 1435 | 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 1436 | 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 1437 | 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 1438 | 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 1439 | 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 1440 | 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 1441 | Deduction adding conjuncts to an equivalence. (Contributed by NM, 8-Sep-2006.) |
⊢ (𝜑 → (𝜓 ↔ 𝜒)) ⇒ ⊢ (𝜑 → ((𝜓 ∧ 𝜃 ∧ 𝜏) ↔ (𝜒 ∧ 𝜃 ∧ 𝜏))) | ||
Theorem | 3anbi2d 1442 | Deduction adding conjuncts to an equivalence. (Contributed by NM, 8-Sep-2006.) |
⊢ (𝜑 → (𝜓 ↔ 𝜒)) ⇒ ⊢ (𝜑 → ((𝜃 ∧ 𝜓 ∧ 𝜏) ↔ (𝜃 ∧ 𝜒 ∧ 𝜏))) | ||
Theorem | 3anbi3d 1443 | Deduction adding conjuncts to an equivalence. (Contributed by NM, 8-Sep-2006.) |
⊢ (𝜑 → (𝜓 ↔ 𝜒)) ⇒ ⊢ (𝜑 → ((𝜃 ∧ 𝜏 ∧ 𝜓) ↔ (𝜃 ∧ 𝜏 ∧ 𝜒))) | ||
Theorem | 3anim123d 1444 | 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 1445 | 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 1446 | Rearrangement of 6 conjuncts. (Contributed by NM, 13-Mar-1995.) |
⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ∧ (𝜃 ∧ 𝜏 ∧ 𝜂)) ↔ ((𝜑 ∧ 𝜃) ∧ (𝜓 ∧ 𝜏) ∧ (𝜒 ∧ 𝜂))) | ||
Theorem | 3an6 1447 | Analogue of an4 655 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 1448 | Analogue of or4 926 for triple conjunction. (Contributed by Scott Fenton, 16-Mar-2011.) |
⊢ (((𝜑 ∨ 𝜓) ∨ (𝜒 ∨ 𝜃) ∨ (𝜏 ∨ 𝜂)) ↔ ((𝜑 ∨ 𝜒 ∨ 𝜏) ∨ (𝜓 ∨ 𝜃 ∨ 𝜂))) | ||
Theorem | mp3an1 1449 | An inference based on modus ponens. (Contributed by NM, 21-Nov-1994.) |
⊢ 𝜑 & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ⇒ ⊢ ((𝜓 ∧ 𝜒) → 𝜃) | ||
Theorem | mp3an2 1450 | An inference based on modus ponens. (Contributed by NM, 21-Nov-1994.) |
⊢ 𝜓 & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ⇒ ⊢ ((𝜑 ∧ 𝜒) → 𝜃) | ||
Theorem | mp3an3 1451 | An inference based on modus ponens. (Contributed by NM, 21-Nov-1994.) |
⊢ 𝜒 & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ⇒ ⊢ ((𝜑 ∧ 𝜓) → 𝜃) | ||
Theorem | mp3an12 1452 | An inference based on modus ponens. (Contributed by NM, 13-Jul-2005.) |
⊢ 𝜑 & ⊢ 𝜓 & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ⇒ ⊢ (𝜒 → 𝜃) | ||
Theorem | mp3an13 1453 | An inference based on modus ponens. (Contributed by NM, 14-Jul-2005.) |
⊢ 𝜑 & ⊢ 𝜒 & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ⇒ ⊢ (𝜓 → 𝜃) | ||
Theorem | mp3an23 1454 | An inference based on modus ponens. (Contributed by NM, 14-Jul-2005.) |
⊢ 𝜓 & ⊢ 𝜒 & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ⇒ ⊢ (𝜑 → 𝜃) | ||
Theorem | mp3an1i 1455 | An inference based on modus ponens. (Contributed by NM, 5-Jul-2005.) |
⊢ 𝜓 & ⊢ (𝜑 → ((𝜓 ∧ 𝜒 ∧ 𝜃) → 𝜏)) ⇒ ⊢ (𝜑 → ((𝜒 ∧ 𝜃) → 𝜏)) | ||
Theorem | mp3anl1 1456 | An inference based on modus ponens. (Contributed by NM, 24-Feb-2005.) |
⊢ 𝜑 & ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ∧ 𝜃) → 𝜏) ⇒ ⊢ (((𝜓 ∧ 𝜒) ∧ 𝜃) → 𝜏) | ||
Theorem | mp3anl2 1457 | An inference based on modus ponens. (Contributed by NM, 24-Feb-2005.) |
⊢ 𝜓 & ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ∧ 𝜃) → 𝜏) ⇒ ⊢ (((𝜑 ∧ 𝜒) ∧ 𝜃) → 𝜏) | ||
Theorem | mp3anl3 1458 | An inference based on modus ponens. (Contributed by NM, 24-Feb-2005.) |
⊢ 𝜒 & ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ∧ 𝜃) → 𝜏) ⇒ ⊢ (((𝜑 ∧ 𝜓) ∧ 𝜃) → 𝜏) | ||
Theorem | mp3anr1 1459 | An inference based on modus ponens. (Contributed by NM, 4-Nov-2006.) |
⊢ 𝜓 & ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒 ∧ 𝜃)) → 𝜏) ⇒ ⊢ ((𝜑 ∧ (𝜒 ∧ 𝜃)) → 𝜏) | ||
Theorem | mp3anr2 1460 | An inference based on modus ponens. (Contributed by NM, 24-Nov-2006.) |
⊢ 𝜒 & ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒 ∧ 𝜃)) → 𝜏) ⇒ ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜃)) → 𝜏) | ||
Theorem | mp3anr3 1461 | An inference based on modus ponens. (Contributed by NM, 19-Oct-2007.) |
⊢ 𝜃 & ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒 ∧ 𝜃)) → 𝜏) ⇒ ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒)) → 𝜏) | ||
Theorem | mp3an 1462 | An inference based on modus ponens. (Contributed by NM, 14-May-1999.) |
⊢ 𝜑 & ⊢ 𝜓 & ⊢ 𝜒 & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ⇒ ⊢ 𝜃 | ||
Theorem | mpd3an3 1463 | An inference based on modus ponens. (Contributed by NM, 8-Nov-2007.) |
⊢ ((𝜑 ∧ 𝜓) → 𝜒) & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ⇒ ⊢ ((𝜑 ∧ 𝜓) → 𝜃) | ||
Theorem | mpd3an23 1464 | An inference based on modus ponens. (Contributed by NM, 4-Dec-2006.) |
⊢ (𝜑 → 𝜓) & ⊢ (𝜑 → 𝜒) & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ⇒ ⊢ (𝜑 → 𝜃) | ||
Theorem | mp3and 1465 | 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 1466 | mp3an 1462 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 1467 | mp3an 1462 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 1468 | mp3an 1462 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 1469 | 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 1470 | Infer implication from a logical equivalence. Similar to biimpa 478. (Contributed by NM, 4-Sep-2005.) |
⊢ ((𝜑 ∧ 𝜓) → (𝜒 ↔ 𝜃)) ⇒ ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) | ||
Theorem | biimp3ar 1471 | Infer implication from a logical equivalence. Similar to biimpar 479. (Contributed by NM, 2-Jan-2009.) |
⊢ ((𝜑 ∧ 𝜓) → (𝜒 ↔ 𝜃)) ⇒ ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜃) → 𝜒) | ||
Theorem | 3anandis 1472 | 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 1473 | 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 1474 | Deduction for elimination by cases. (Contributed by NM, 22-Apr-1994.) |
⊢ (𝜑 → ¬ 𝜒) & ⊢ (𝜑 → ¬ 𝜃) & ⊢ (𝜑 → (𝜓 ∨ 𝜒 ∨ 𝜃)) ⇒ ⊢ (𝜑 → 𝜓) | ||
Theorem | 3ecase 1475 | Inference for elimination by cases. (Contributed by NM, 13-Jul-2005.) |
⊢ (¬ 𝜑 → 𝜃) & ⊢ (¬ 𝜓 → 𝜃) & ⊢ (¬ 𝜒 → 𝜃) & ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃) ⇒ ⊢ 𝜃 | ||
Theorem | 3bior1fd 1476 | A disjunction is equivalent to a threefold disjunction with single falsehood, analogous to biorf 936. (Contributed by Alexander van der Vekens, 8-Sep-2017.) |
⊢ (𝜑 → ¬ 𝜃) ⇒ ⊢ (𝜑 → ((𝜒 ∨ 𝜓) ↔ (𝜃 ∨ 𝜒 ∨ 𝜓))) | ||
Theorem | 3bior1fand 1477 | 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 1478 | A wff is equivalent to its threefold disjunction with double falsehood, analogous to biorf 936. (Contributed by Alexander van der Vekens, 8-Sep-2017.) |
⊢ (𝜑 → ¬ 𝜃) & ⊢ (𝜑 → ¬ 𝜒) ⇒ ⊢ (𝜑 → (𝜓 ↔ (𝜃 ∨ 𝜒 ∨ 𝜓))) | ||
Theorem | 3biant1d 1479 | A conjunction is equivalent to a threefold conjunction with single truth, analogous to biantrud 533. (Contributed by Alexander van der Vekens, 26-Sep-2017.) |
⊢ (𝜑 → 𝜃) ⇒ ⊢ (𝜑 → ((𝜒 ∧ 𝜓) ↔ (𝜃 ∧ 𝜒 ∧ 𝜓))) | ||
Theorem | intn3an1d 1480 | 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 1481 | 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 1482 | 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 1483 | 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 1484 | Rearrange a 9-fold conjunction. (Contributed by Thierry Arnoux, 14-Apr-2019.) (Proof shortened by Wolf Lammen, 21-Apr-2024.) |
⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ∧ (𝜃 ∧ 𝜏 ∧ 𝜂) ∧ (𝜁 ∧ 𝜎 ∧ 𝜌)) ↔ ((𝜑 ∧ 𝜏 ∧ 𝜌) ∧ ((𝜓 ∧ 𝜃) ∧ (𝜂 ∧ 𝜎) ∧ (𝜒 ∧ 𝜁)))) | ||
Theorem | an33reanOLD 1485 | Obsolete version of an33rean 1484 as of 21-Apr-2024. (Contributed by Thierry Arnoux, 14-Apr-2019.) (Proof modification is discouraged.) (New usage is discouraged.) |
⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ∧ (𝜃 ∧ 𝜏 ∧ 𝜂) ∧ (𝜁 ∧ 𝜎 ∧ 𝜌)) ↔ ((𝜑 ∧ 𝜏 ∧ 𝜌) ∧ ((𝜓 ∧ 𝜃) ∧ (𝜂 ∧ 𝜎) ∧ (𝜒 ∧ 𝜁)))) | ||
Theorem | 3orel2 1486 | Partial elimination of a triple disjunction by denial of a disjunct. (Contributed by Scott Fenton, 26-Mar-2011.) (Proof shortened by Andrew Salmon, 25-May-2011.) |
⊢ (¬ 𝜓 → ((𝜑 ∨ 𝜓 ∨ 𝜒) → (𝜑 ∨ 𝜒))) | ||
Theorem | 3orel3 1487 | Partial elimination of a triple disjunction by denial of a disjunct. (Contributed by Scott Fenton, 26-Mar-2011.) |
⊢ (¬ 𝜒 → ((𝜑 ∨ 𝜓 ∨ 𝜒) → (𝜑 ∨ 𝜓))) | ||
Theorem | 3orel13 1488 | Elimination of two disjuncts in a triple disjunction. (Contributed by Scott Fenton, 9-Jun-2011.) |
⊢ ((¬ 𝜑 ∧ ¬ 𝜒) → ((𝜑 ∨ 𝜓 ∨ 𝜒) → 𝜓)) | ||
Theorem | 3pm3.2ni 1489 | Triple negated disjunction introduction. (Contributed by Scott Fenton, 20-Apr-2011.) |
⊢ ¬ 𝜑 & ⊢ ¬ 𝜓 & ⊢ ¬ 𝜒 ⇒ ⊢ ¬ (𝜑 ∨ 𝜓 ∨ 𝜒) | ||
Syntax | wnan 1490 | Extend wff definition to include alternative denial ("nand"). |
wff (𝜑 ⊼ 𝜓) | ||
Definition | df-nan 1491 | Define incompatibility, or alternative denial ("not-and" or "nand"). See dfnan2 1493 for an alternative. 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 1545) and the constant false ⊥ (df-fal 1555), we will be able to prove these truth table values: ((⊤ ⊼ ⊤) ↔ ⊥) (trunantru 1583), ((⊤ ⊼ ⊥) ↔ ⊤) (trunanfal 1584), ((⊥ ⊼ ⊤) ↔ ⊤) (falnantru 1585), and ((⊥ ⊼ ⊥) ↔ ⊤) (falnanfal 1586). Contrast with ∧ (df-an 398), ∨ (df-or 847), → (wi 4), and ⊻ (df-xor 1511). (Contributed by Jeff Hoffman, 19-Nov-2007.) |
⊢ ((𝜑 ⊼ 𝜓) ↔ ¬ (𝜑 ∧ 𝜓)) | ||
Theorem | nanan 1492 | Conjunction in terms of alternative denial. (Contributed by Mario Carneiro, 9-May-2015.) |
⊢ ((𝜑 ∧ 𝜓) ↔ ¬ (𝜑 ⊼ 𝜓)) | ||
Theorem | dfnan2 1493 | Alternative denial in terms of our primitive connectives (implication and negation). (Contributed by WL, 26-Jun-2020.) |
⊢ ((𝜑 ⊼ 𝜓) ↔ (𝜑 → ¬ 𝜓)) | ||
Theorem | nanor 1494 | 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 1495 | Alternative denial is commutative. Remark: alternative denial is not associative, see nanass 1509. (Contributed by Mario Carneiro, 9-May-2015.) (Proof shortened by Wolf Lammen, 26-Jun-2020.) |
⊢ ((𝜑 ⊼ 𝜓) ↔ (𝜓 ⊼ 𝜑)) | ||
Theorem | nannan 1496 | 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 1497 | Implication in terms of alternative denial. (Contributed by Jeff Hoffman, 19-Nov-2007.) |
⊢ ((𝜑 → 𝜓) ↔ (𝜑 ⊼ (𝜓 ⊼ 𝜓))) | ||
Theorem | nannot 1498 | 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 1499 | 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 1500 | 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))) |
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