P. 15 of Theorem List - Metamath Proof Explorer

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

Type	Label	Description
Statement
Theorem	syl332anc 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	syl333anc 1402	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 1403	A syllogism inference. (Contributed by NM, 22-Aug-1995.)
⊢ (𝜑 ↔ 𝜓)    &   ⊢ ((𝜓 ∧ 𝜒 ∧ 𝜃) → 𝜏)    ⇒   ⊢ ((𝜑 ∧ 𝜒 ∧ 𝜃) → 𝜏)
Theorem	syl3an2b 1404	A syllogism inference. (Contributed by NM, 22-Aug-1995.)
⊢ (𝜑 ↔ 𝜒)    &   ⊢ ((𝜓 ∧ 𝜒 ∧ 𝜃) → 𝜏)    ⇒   ⊢ ((𝜓 ∧ 𝜑 ∧ 𝜃) → 𝜏)
Theorem	syl3an3b 1405	A syllogism inference. (Contributed by NM, 22-Aug-1995.)
⊢ (𝜑 ↔ 𝜃)    &   ⊢ ((𝜓 ∧ 𝜒 ∧ 𝜃) → 𝜏)    ⇒   ⊢ ((𝜓 ∧ 𝜒 ∧ 𝜑) → 𝜏)
Theorem	syl3an1br 1406	A syllogism inference. (Contributed by NM, 22-Aug-1995.)
⊢ (𝜓 ↔ 𝜑)    &   ⊢ ((𝜓 ∧ 𝜒 ∧ 𝜃) → 𝜏)    ⇒   ⊢ ((𝜑 ∧ 𝜒 ∧ 𝜃) → 𝜏)
Theorem	syl3an2br 1407	A syllogism inference. (Contributed by NM, 22-Aug-1995.)
⊢ (𝜒 ↔ 𝜑)    &   ⊢ ((𝜓 ∧ 𝜒 ∧ 𝜃) → 𝜏)    ⇒   ⊢ ((𝜓 ∧ 𝜑 ∧ 𝜃) → 𝜏)
Theorem	syl3an3br 1408	A syllogism inference. (Contributed by NM, 22-Aug-1995.)
⊢ (𝜃 ↔ 𝜑)    &   ⊢ ((𝜓 ∧ 𝜒 ∧ 𝜃) → 𝜏)    ⇒   ⊢ ((𝜓 ∧ 𝜒 ∧ 𝜑) → 𝜏)
Theorem	syld3an3 1409	A syllogism inference. (Contributed by NM, 20-May-2007.)
⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)    &   ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜃) → 𝜏)    ⇒   ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜏)
Theorem	syld3an1 1410	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 1411	A syllogism inference. (Contributed by NM, 20-May-2007.)
⊢ ((𝜑 ∧ 𝜒 ∧ 𝜃) → 𝜓)    &   ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜃) → 𝜏)    ⇒   ⊢ ((𝜑 ∧ 𝜒 ∧ 𝜃) → 𝜏)
Theorem	syl3anl1 1412	A syllogism inference. (Contributed by NM, 24-Feb-2005.)
⊢ (𝜑 → 𝜓)    &   ⊢ (((𝜓 ∧ 𝜒 ∧ 𝜃) ∧ 𝜏) → 𝜂)    ⇒   ⊢ (((𝜑 ∧ 𝜒 ∧ 𝜃) ∧ 𝜏) → 𝜂)
Theorem	syl3anl2 1413	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 1414	A syllogism inference. (Contributed by NM, 24-Feb-2005.)
⊢ (𝜑 → 𝜃)    &   ⊢ (((𝜓 ∧ 𝜒 ∧ 𝜃) ∧ 𝜏) → 𝜂)    ⇒   ⊢ (((𝜓 ∧ 𝜒 ∧ 𝜑) ∧ 𝜏) → 𝜂)
Theorem	syl3anl 1415	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 1416	A syllogism inference. (Contributed by NM, 31-Jul-2007.)
⊢ (𝜑 → 𝜓)    &   ⊢ ((𝜒 ∧ (𝜓 ∧ 𝜃 ∧ 𝜏)) → 𝜂)    ⇒   ⊢ ((𝜒 ∧ (𝜑 ∧ 𝜃 ∧ 𝜏)) → 𝜂)
Theorem	syl3anr2 1417	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 1418	A syllogism inference. (Contributed by NM, 23-Aug-2007.)
⊢ (𝜑 → 𝜏)    &   ⊢ ((𝜒 ∧ (𝜓 ∧ 𝜃 ∧ 𝜏)) → 𝜂)    ⇒   ⊢ ((𝜒 ∧ (𝜓 ∧ 𝜃 ∧ 𝜑)) → 𝜂)
Theorem	3anidm12 1419	Inference from idempotent law for conjunction. (Contributed by NM, 7-Mar-2008.)
⊢ ((𝜑 ∧ 𝜑 ∧ 𝜓) → 𝜒)    ⇒   ⊢ ((𝜑 ∧ 𝜓) → 𝜒)
Theorem	3anidm13 1420	Inference from idempotent law for conjunction. (Contributed by NM, 7-Mar-2008.)
⊢ ((𝜑 ∧ 𝜓 ∧ 𝜑) → 𝜒)    ⇒   ⊢ ((𝜑 ∧ 𝜓) → 𝜒)
Theorem	3anidm23 1421	Inference from idempotent law for conjunction. (Contributed by NM, 1-Feb-2007.)
⊢ ((𝜑 ∧ 𝜓 ∧ 𝜓) → 𝜒)    ⇒   ⊢ ((𝜑 ∧ 𝜓) → 𝜒)
Theorem	syl2an3an 1422	syl3an 1160 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 1423	Deduction related to syl3an 1160 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 1424	Infer implication from triple disjunction. (Contributed by NM, 26-Sep-2006.)
⊢ (𝜑 ∨ 𝜓 ∨ 𝜒)    ⇒   ⊢ ((¬ 𝜑 ∧ ¬ 𝜓) → 𝜒)
Theorem	3jao 1425	Disjunction of three antecedents. (Contributed by NM, 8-Apr-1994.)
⊢ (((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)) → ((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓))
Theorem	3jaob 1426	Disjunction of three antecedents. (Contributed by NM, 13-Sep-2011.) (Proof shortened by Hongxiu Chen, 29-Jun-2025.)
⊢ (((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓) ↔ ((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)))
Theorem	3jaobOLD 1427	Obsolete version of 3jaob 1426 as of 29-Jun-2025. (Contributed by NM, 13-Sep-2011.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ (((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	3jaao 1433	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 1434	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 1435	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 1436	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 1437	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 1438	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 1439	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 1440	Deduction adding conjuncts to an equivalence. (Contributed by NM, 8-Sep-2006.)
⊢ (𝜑 → (𝜓 ↔ 𝜒))    ⇒   ⊢ (𝜑 → ((𝜓 ∧ 𝜃 ∧ 𝜏) ↔ (𝜒 ∧ 𝜃 ∧ 𝜏)))
Theorem	3anbi2d 1441	Deduction adding conjuncts to an equivalence. (Contributed by NM, 8-Sep-2006.)
⊢ (𝜑 → (𝜓 ↔ 𝜒))    ⇒   ⊢ (𝜑 → ((𝜃 ∧ 𝜓 ∧ 𝜏) ↔ (𝜃 ∧ 𝜒 ∧ 𝜏)))
Theorem	3anbi3d 1442	Deduction adding conjuncts to an equivalence. (Contributed by NM, 8-Sep-2006.)
⊢ (𝜑 → (𝜓 ↔ 𝜒))    ⇒   ⊢ (𝜑 → ((𝜃 ∧ 𝜏 ∧ 𝜓) ↔ (𝜃 ∧ 𝜏 ∧ 𝜒)))
Theorem	3anim123d 1443	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 1444	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 1445	Rearrangement of 6 conjuncts. (Contributed by NM, 13-Mar-1995.)
⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ∧ (𝜃 ∧ 𝜏 ∧ 𝜂)) ↔ ((𝜑 ∧ 𝜃) ∧ (𝜓 ∧ 𝜏) ∧ (𝜒 ∧ 𝜂)))
Theorem	3an6 1446	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 1447	Analogue of or4 925 for triple conjunction. (Contributed by Scott Fenton, 16-Mar-2011.)
⊢ (((𝜑 ∨ 𝜓) ∨ (𝜒 ∨ 𝜃) ∨ (𝜏 ∨ 𝜂)) ↔ ((𝜑 ∨ 𝜒 ∨ 𝜏) ∨ (𝜓 ∨ 𝜃 ∨ 𝜂)))
Theorem	mp3an1 1448	An inference based on modus ponens. (Contributed by NM, 21-Nov-1994.)
⊢ 𝜑    &   ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)    ⇒   ⊢ ((𝜓 ∧ 𝜒) → 𝜃)
Theorem	mp3an2 1449	An inference based on modus ponens. (Contributed by NM, 21-Nov-1994.)
⊢ 𝜓    &   ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)    ⇒   ⊢ ((𝜑 ∧ 𝜒) → 𝜃)
Theorem	mp3an3 1450	An inference based on modus ponens. (Contributed by NM, 21-Nov-1994.)
⊢ 𝜒    &   ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)    ⇒   ⊢ ((𝜑 ∧ 𝜓) → 𝜃)
Theorem	mp3an12 1451	An inference based on modus ponens. (Contributed by NM, 13-Jul-2005.)
⊢ 𝜑    &   ⊢ 𝜓    &   ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)    ⇒   ⊢ (𝜒 → 𝜃)
Theorem	mp3an13 1452	An inference based on modus ponens. (Contributed by NM, 14-Jul-2005.)
⊢ 𝜑    &   ⊢ 𝜒    &   ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)    ⇒   ⊢ (𝜓 → 𝜃)
Theorem	mp3an23 1453	An inference based on modus ponens. (Contributed by NM, 14-Jul-2005.)
⊢ 𝜓    &   ⊢ 𝜒    &   ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)    ⇒   ⊢ (𝜑 → 𝜃)
Theorem	mp3an1i 1454	An inference based on modus ponens. (Contributed by NM, 5-Jul-2005.)
⊢ 𝜓    &   ⊢ (𝜑 → ((𝜓 ∧ 𝜒 ∧ 𝜃) → 𝜏))    ⇒   ⊢ (𝜑 → ((𝜒 ∧ 𝜃) → 𝜏))
Theorem	mp3anl1 1455	An inference based on modus ponens. (Contributed by NM, 24-Feb-2005.)
⊢ 𝜑    &   ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ∧ 𝜃) → 𝜏)    ⇒   ⊢ (((𝜓 ∧ 𝜒) ∧ 𝜃) → 𝜏)
Theorem	mp3anl2 1456	An inference based on modus ponens. (Contributed by NM, 24-Feb-2005.)
⊢ 𝜓    &   ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ∧ 𝜃) → 𝜏)    ⇒   ⊢ (((𝜑 ∧ 𝜒) ∧ 𝜃) → 𝜏)
Theorem	mp3anl3 1457	An inference based on modus ponens. (Contributed by NM, 24-Feb-2005.)
⊢ 𝜒    &   ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ∧ 𝜃) → 𝜏)    ⇒   ⊢ (((𝜑 ∧ 𝜓) ∧ 𝜃) → 𝜏)
Theorem	mp3anr1 1458	An inference based on modus ponens. (Contributed by NM, 4-Nov-2006.)
⊢ 𝜓    &   ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒 ∧ 𝜃)) → 𝜏)    ⇒   ⊢ ((𝜑 ∧ (𝜒 ∧ 𝜃)) → 𝜏)
Theorem	mp3anr2 1459	An inference based on modus ponens. (Contributed by NM, 24-Nov-2006.)
⊢ 𝜒    &   ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒 ∧ 𝜃)) → 𝜏)    ⇒   ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜃)) → 𝜏)
Theorem	mp3anr3 1460	An inference based on modus ponens. (Contributed by NM, 19-Oct-2007.)
⊢ 𝜃    &   ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒 ∧ 𝜃)) → 𝜏)    ⇒   ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒)) → 𝜏)
Theorem	mp3an 1461	An inference based on modus ponens. (Contributed by NM, 14-May-1999.)
⊢ 𝜑    &   ⊢ 𝜓    &   ⊢ 𝜒    &   ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)    ⇒   ⊢ 𝜃
Theorem	mpd3an3 1462	An inference based on modus ponens. (Contributed by NM, 8-Nov-2007.)
⊢ ((𝜑 ∧ 𝜓) → 𝜒)    &   ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)    ⇒   ⊢ ((𝜑 ∧ 𝜓) → 𝜃)
Theorem	mpd3an23 1463	An inference based on modus ponens. (Contributed by NM, 4-Dec-2006.)
⊢ (𝜑 → 𝜓)    &   ⊢ (𝜑 → 𝜒)    &   ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)    ⇒   ⊢ (𝜑 → 𝜃)
Theorem	mp3and 1464	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 1465	mp3an 1461 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 1466	mp3an 1461 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 1467	mp3an 1461 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 1468	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 1469	Infer implication from a logical equivalence. Similar to biimpa 476. (Contributed by NM, 4-Sep-2005.)
⊢ ((𝜑 ∧ 𝜓) → (𝜒 ↔ 𝜃))    ⇒   ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)
Theorem	biimp3ar 1470	Infer implication from a logical equivalence. Similar to biimpar 477. (Contributed by NM, 2-Jan-2009.)
⊢ ((𝜑 ∧ 𝜓) → (𝜒 ↔ 𝜃))    ⇒   ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜃) → 𝜒)
Theorem	3anandis 1471	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 1472	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 1473	Deduction for elimination by cases. (Contributed by NM, 22-Apr-1994.)
⊢ (𝜑 → ¬ 𝜒)    &   ⊢ (𝜑 → ¬ 𝜃)    &   ⊢ (𝜑 → (𝜓 ∨ 𝜒 ∨ 𝜃))    ⇒   ⊢ (𝜑 → 𝜓)
Theorem	3ecase 1474	Inference for elimination by cases. (Contributed by NM, 13-Jul-2005.)
⊢ (¬ 𝜑 → 𝜃)    &   ⊢ (¬ 𝜓 → 𝜃)    &   ⊢ (¬ 𝜒 → 𝜃)    &   ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → 𝜃)    ⇒   ⊢ 𝜃
Theorem	3bior1fd 1475	A disjunction is equivalent to a threefold disjunction with single falsehood, analogous to biorf 935. (Contributed by Alexander van der Vekens, 8-Sep-2017.)
⊢ (𝜑 → ¬ 𝜃)    ⇒   ⊢ (𝜑 → ((𝜒 ∨ 𝜓) ↔ (𝜃 ∨ 𝜒 ∨ 𝜓)))
Theorem	3bior1fand 1476	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 1477	A wff is equivalent to its threefold disjunction with double falsehood, analogous to biorf 935. (Contributed by Alexander van der Vekens, 8-Sep-2017.)
⊢ (𝜑 → ¬ 𝜃)    &   ⊢ (𝜑 → ¬ 𝜒)    ⇒   ⊢ (𝜑 → (𝜓 ↔ (𝜃 ∨ 𝜒 ∨ 𝜓)))
Theorem	3biant1d 1478	A conjunction is equivalent to a threefold conjunction with single truth, analogous to biantrud 531. (Contributed by Alexander van der Vekens, 26-Sep-2017.)
⊢ (𝜑 → 𝜃)    ⇒   ⊢ (𝜑 → ((𝜒 ∧ 𝜓) ↔ (𝜃 ∧ 𝜒 ∧ 𝜓)))
Theorem	intn3an1d 1479	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 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 → ¬ (𝜒 ∧ 𝜓 ∧ 𝜃))
Theorem	intn3an3d 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 ∧ 𝜃 ∧ 𝜓))
Theorem	an3andi 1482	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 1483	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	3orel2 1484	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 1485	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 1486	Elimination of two disjuncts in a triple disjunction. (Contributed by Scott Fenton, 9-Jun-2011.)
⊢ ((¬ 𝜑 ∧ ¬ 𝜒) → ((𝜑 ∨ 𝜓 ∨ 𝜒) → 𝜓))
Theorem	3pm3.2ni 1487	Triple negated disjunction introduction. (Contributed by Scott Fenton, 20-Apr-2011.)
⊢ ¬ 𝜑    &   ⊢ ¬ 𝜓    &   ⊢ ¬ 𝜒    ⇒   ⊢ ¬ (𝜑 ∨ 𝜓 ∨ 𝜒)
1.2.12  Logical "nand" (Sheffer stroke)
Syntax	wnan 1488	Extend wff definition to include alternative denial ("nand").
wff (𝜑 ⊼ 𝜓)
Definition	df-nan 1489	Define incompatibility, or alternative denial ("not-and" or "nand"). See dfnan2 1491 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 1540) and the constant false ⊥ (df-fal 1550), we will be able to prove these truth table values: ((⊤ ⊼ ⊤) ↔ ⊥) (trunantru 1578), ((⊤ ⊼ ⊥) ↔ ⊤) (trunanfal 1579), ((⊥ ⊼ ⊤) ↔ ⊤) (falnantru 1580), and ((⊥ ⊼ ⊥) ↔ ⊤) (falnanfal 1581). Contrast with ∧ (df-an 396), ∨ (df-or 847), → (wi 4), and ⊻ (df-xor 1509). (Contributed by Jeff Hoffman, 19-Nov-2007.)
⊢ ((𝜑 ⊼ 𝜓) ↔ ¬ (𝜑 ∧ 𝜓))
Theorem	nanan 1490	Conjunction in terms of alternative denial. (Contributed by Mario Carneiro, 9-May-2015.)
⊢ ((𝜑 ∧ 𝜓) ↔ ¬ (𝜑 ⊼ 𝜓))
Theorem	dfnan2 1491	Alternative denial in terms of our primitive connectives (implication and negation). (Contributed by WL, 26-Jun-2020.)
⊢ ((𝜑 ⊼ 𝜓) ↔ (𝜑 → ¬ 𝜓))
Theorem	nanor 1492	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 1493	Alternative denial is commutative. Remark: alternative denial is not associative, see nanass 1507. (Contributed by Mario Carneiro, 9-May-2015.) (Proof shortened by Wolf Lammen, 26-Jun-2020.)
⊢ ((𝜑 ⊼ 𝜓) ↔ (𝜓 ⊼ 𝜑))
Theorem	nannan 1494	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 1495	Implication in terms of alternative denial. (Contributed by Jeff Hoffman, 19-Nov-2007.)
⊢ ((𝜑 → 𝜓) ↔ (𝜑 ⊼ (𝜓 ⊼ 𝜓)))
Theorem	nannot 1496	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 1497	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 1498	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 1499	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 1500	Join two logical equivalences with anti-conjunction. (Contributed by SF, 2-Jan-2018.)
⊢ (((𝜑 ↔ 𝜓) ∧ (𝜒 ↔ 𝜃)) → ((𝜑 ⊼ 𝜒) ↔ (𝜓 ⊼ 𝜃)))