P. 124 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 12301-12400   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	1p1e2 12301	1 + 1 = 2. (Contributed by NM, 1-Apr-2008.)
⊢ (1 + 1) = 2
Theorem	2m1e1 12302	2 - 1 = 1. The result is on the right-hand-side to be consistent with similar proofs like 4p4e8 12331. (Contributed by David A. Wheeler, 4-Jan-2017.)
⊢ (2 − 1) = 1
Theorem	1e2m1 12303	1 = 2 - 1. (Contributed by David A. Wheeler, 8-Dec-2018.)
⊢ 1 = (2 − 1)
Theorem	3m1e2 12304	3 - 1 = 2. (Contributed by FL, 17-Oct-2010.) (Revised by NM, 10-Dec-2017.) (Proof shortened by AV, 6-Sep-2021.)
⊢ (3 − 1) = 2
Theorem	4m1e3 12305	4 - 1 = 3. (Contributed by AV, 8-Feb-2021.) (Proof shortened by AV, 6-Sep-2021.)
⊢ (4 − 1) = 3
Theorem	5m1e4 12306	5 - 1 = 4. (Contributed by AV, 6-Sep-2021.)
⊢ (5 − 1) = 4
Theorem	6m1e5 12307	6 - 1 = 5. (Contributed by AV, 6-Sep-2021.)
⊢ (6 − 1) = 5
Theorem	7m1e6 12308	7 - 1 = 6. (Contributed by AV, 6-Sep-2021.)
⊢ (7 − 1) = 6
Theorem	8m1e7 12309	8 - 1 = 7. (Contributed by AV, 6-Sep-2021.)
⊢ (8 − 1) = 7
Theorem	9m1e8 12310	9 - 1 = 8. (Contributed by AV, 6-Sep-2021.)
⊢ (9 − 1) = 8
Theorem	2p2e4 12311	Two plus two equals four. For more information, see "2+2=4 Trivia" on the Metamath Proof Explorer Home Page: mmset.html#trivia. This proof is simple, but it depends on many other proof steps because 2 and 4 are complex numbers and thus it depends on our construction of complex numbers. The proof o2p2e4 8476 is similar but proves 2 + 2 = 4 using ordinal natural numbers (finite integers starting at 0), so that proof depends on fewer intermediate steps. (Contributed by NM, 27-May-1999.)
⊢ (2 + 2) = 4
Theorem	2times 12312	Two times a number. (Contributed by NM, 10-Oct-2004.) (Revised by Mario Carneiro, 27-May-2016.) (Proof shortened by AV, 26-Feb-2020.)
⊢ (𝐴 ∈ ℂ → (2 · 𝐴) = (𝐴 + 𝐴))
Theorem	times2 12313	A number times 2. (Contributed by NM, 16-Oct-2007.)
⊢ (𝐴 ∈ ℂ → (𝐴 · 2) = (𝐴 + 𝐴))
Theorem	2timesi 12314	Two times a number. (Contributed by NM, 1-Aug-1999.)
⊢ 𝐴 ∈ ℂ    ⇒   ⊢ (2 · 𝐴) = (𝐴 + 𝐴)
Theorem	times2i 12315	A number times 2. (Contributed by NM, 11-May-2004.)
⊢ 𝐴 ∈ ℂ    ⇒   ⊢ (𝐴 · 2) = (𝐴 + 𝐴)
Theorem	2txmxeqx 12316	Two times a complex number minus the number itself results in the number itself. (Contributed by Alexander van der Vekens, 8-Jun-2018.)
⊢ (𝑋 ∈ ℂ → ((2 · 𝑋) − 𝑋) = 𝑋)
Theorem	2div2e1 12317	2 divided by 2 is 1. (Contributed by David A. Wheeler, 8-Dec-2018.)
⊢ (2 / 2) = 1
Theorem	2p1e3 12318	2 + 1 = 3. (Contributed by Mario Carneiro, 18-Apr-2015.)
⊢ (2 + 1) = 3
Theorem	1p2e3 12319	1 + 2 = 3. For a shorter proof using addcomli 11338, see 1p2e3ALT 12320. (Contributed by David A. Wheeler, 8-Dec-2018.) Reduce dependencies on axioms. (Revised by Steven Nguyen, 12-Dec-2022.)
⊢ (1 + 2) = 3
Theorem	1p2e3ALT 12320	Alternate proof of 1p2e3 12319, shorter but using more axioms. (Contributed by David A. Wheeler, 8-Dec-2018.) (New usage is discouraged.) (Proof modification is discouraged.)
⊢ (1 + 2) = 3
Theorem	3p1e4 12321	3 + 1 = 4. (Contributed by Mario Carneiro, 18-Apr-2015.)
⊢ (3 + 1) = 4
Theorem	4p1e5 12322	4 + 1 = 5. (Contributed by Mario Carneiro, 18-Apr-2015.)
⊢ (4 + 1) = 5
Theorem	5p1e6 12323	5 + 1 = 6. (Contributed by Mario Carneiro, 18-Apr-2015.)
⊢ (5 + 1) = 6
Theorem	6p1e7 12324	6 + 1 = 7. (Contributed by Mario Carneiro, 18-Apr-2015.)
⊢ (6 + 1) = 7
Theorem	7p1e8 12325	7 + 1 = 8. (Contributed by Mario Carneiro, 18-Apr-2015.)
⊢ (7 + 1) = 8
Theorem	8p1e9 12326	8 + 1 = 9. (Contributed by Mario Carneiro, 18-Apr-2015.)
⊢ (8 + 1) = 9
Theorem	3p2e5 12327	3 + 2 = 5. (Contributed by NM, 11-May-2004.)
⊢ (3 + 2) = 5
Theorem	3p3e6 12328	3 + 3 = 6. (Contributed by NM, 11-May-2004.)
⊢ (3 + 3) = 6
Theorem	4p2e6 12329	4 + 2 = 6. (Contributed by NM, 11-May-2004.)
⊢ (4 + 2) = 6
Theorem	4p3e7 12330	4 + 3 = 7. (Contributed by NM, 11-May-2004.)
⊢ (4 + 3) = 7
Theorem	4p4e8 12331	4 + 4 = 8. (Contributed by NM, 11-May-2004.)
⊢ (4 + 4) = 8
Theorem	5p2e7 12332	5 + 2 = 7. (Contributed by NM, 11-May-2004.)
⊢ (5 + 2) = 7
Theorem	5p3e8 12333	5 + 3 = 8. (Contributed by NM, 11-May-2004.)
⊢ (5 + 3) = 8
Theorem	5p4e9 12334	5 + 4 = 9. (Contributed by NM, 11-May-2004.)
⊢ (5 + 4) = 9
Theorem	6p2e8 12335	6 + 2 = 8. (Contributed by NM, 11-May-2004.)
⊢ (6 + 2) = 8
Theorem	6p3e9 12336	6 + 3 = 9. (Contributed by NM, 11-May-2004.)
⊢ (6 + 3) = 9
Theorem	7p2e9 12337	7 + 2 = 9. (Contributed by NM, 11-May-2004.)
⊢ (7 + 2) = 9
Theorem	1t1e1 12338	1 times 1 equals 1. (Contributed by David A. Wheeler, 7-Jul-2016.)
⊢ (1 · 1) = 1
Theorem	2t1e2 12339	2 times 1 equals 2. (Contributed by David A. Wheeler, 6-Dec-2018.)
⊢ (2 · 1) = 2
Theorem	2t2e4 12340	2 times 2 equals 4. (Contributed by NM, 1-Aug-1999.)
⊢ (2 · 2) = 4
Theorem	3t1e3 12341	3 times 1 equals 3. (Contributed by David A. Wheeler, 8-Dec-2018.)
⊢ (3 · 1) = 3
Theorem	3t2e6 12342	3 times 2 equals 6. (Contributed by NM, 2-Aug-2004.)
⊢ (3 · 2) = 6
Theorem	3t3e9 12343	3 times 3 equals 9. (Contributed by NM, 11-May-2004.)
⊢ (3 · 3) = 9
Theorem	4t2e8 12344	4 times 2 equals 8. (Contributed by NM, 2-Aug-2004.)
⊢ (4 · 2) = 8
Theorem	2t0e0 12345	2 times 0 equals 0. (Contributed by David A. Wheeler, 8-Dec-2018.)
⊢ (2 · 0) = 0
Theorem	4div2e2 12346	One half of four is two. (Contributed by NM, 3-Sep-1999.)
⊢ (4 / 2) = 2
Theorem	1lt2 12347	1 is less than 2. (Contributed by NM, 24-Feb-2005.)
⊢ 1 < 2
Theorem	2lt3 12348	2 is less than 3. (Contributed by NM, 26-Sep-2010.)
⊢ 2 < 3
Theorem	1lt3 12349	1 is less than 3. (Contributed by NM, 26-Sep-2010.)
⊢ 1 < 3
Theorem	3lt4 12350	3 is less than 4. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 3 < 4
Theorem	2lt4 12351	2 is less than 4. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 2 < 4
Theorem	1lt4 12352	1 is less than 4. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 1 < 4
Theorem	4lt5 12353	4 is less than 5. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 4 < 5
Theorem	3lt5 12354	3 is less than 5. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 3 < 5
Theorem	2lt5 12355	2 is less than 5. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 2 < 5
Theorem	1lt5 12356	1 is less than 5. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 1 < 5
Theorem	5lt6 12357	5 is less than 6. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 5 < 6
Theorem	4lt6 12358	4 is less than 6. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 4 < 6
Theorem	3lt6 12359	3 is less than 6. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 3 < 6
Theorem	2lt6 12360	2 is less than 6. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 2 < 6
Theorem	1lt6 12361	1 is less than 6. (Contributed by NM, 19-Oct-2012.)
⊢ 1 < 6
Theorem	6lt7 12362	6 is less than 7. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 6 < 7
Theorem	5lt7 12363	5 is less than 7. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 5 < 7
Theorem	4lt7 12364	4 is less than 7. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 4 < 7
Theorem	3lt7 12365	3 is less than 7. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 3 < 7
Theorem	2lt7 12366	2 is less than 7. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 2 < 7
Theorem	1lt7 12367	1 is less than 7. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 1 < 7
Theorem	7lt8 12368	7 is less than 8. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 7 < 8
Theorem	6lt8 12369	6 is less than 8. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 6 < 8
Theorem	5lt8 12370	5 is less than 8. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 5 < 8
Theorem	4lt8 12371	4 is less than 8. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 4 < 8
Theorem	3lt8 12372	3 is less than 8. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 3 < 8
Theorem	2lt8 12373	2 is less than 8. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 2 < 8
Theorem	1lt8 12374	1 is less than 8. (Contributed by Mario Carneiro, 15-Sep-2013.)
⊢ 1 < 8
Theorem	8lt9 12375	8 is less than 9. (Contributed by Mario Carneiro, 19-Feb-2014.)
⊢ 8 < 9
Theorem	7lt9 12376	7 is less than 9. (Contributed by Mario Carneiro, 9-Mar-2015.)
⊢ 7 < 9
Theorem	6lt9 12377	6 is less than 9. (Contributed by Mario Carneiro, 9-Mar-2015.)
⊢ 6 < 9
Theorem	5lt9 12378	5 is less than 9. (Contributed by Mario Carneiro, 9-Mar-2015.)
⊢ 5 < 9
Theorem	4lt9 12379	4 is less than 9. (Contributed by Mario Carneiro, 9-Mar-2015.)
⊢ 4 < 9
Theorem	3lt9 12380	3 is less than 9. (Contributed by Mario Carneiro, 9-Mar-2015.)
⊢ 3 < 9
Theorem	2lt9 12381	2 is less than 9. (Contributed by Mario Carneiro, 9-Mar-2015.)
⊢ 2 < 9
Theorem	1lt9 12382	1 is less than 9. (Contributed by NM, 19-Oct-2012.) (Revised by Mario Carneiro, 9-Mar-2015.)
⊢ 1 < 9
Theorem	0ne2 12383	0 is not equal to 2. (Contributed by David A. Wheeler, 8-Dec-2018.)
⊢ 0 ≠ 2
Theorem	1ne2 12384	1 is not equal to 2. (Contributed by NM, 19-Oct-2012.)
⊢ 1 ≠ 2
Theorem	1le2 12385	1 is less than or equal to 2. (Contributed by David A. Wheeler, 8-Dec-2018.)
⊢ 1 ≤ 2
Theorem	2cnne0 12386	2 is a nonzero complex number. (Contributed by David A. Wheeler, 7-Dec-2018.)
⊢ (2 ∈ ℂ ∧ 2 ≠ 0)
Theorem	2rene0 12387	2 is a nonzero real number. (Contributed by David A. Wheeler, 8-Dec-2018.)
⊢ (2 ∈ ℝ ∧ 2 ≠ 0)
Theorem	1le3 12388	1 is less than or equal to 3. (Contributed by David A. Wheeler, 8-Dec-2018.)
⊢ 1 ≤ 3
Theorem	neg1mulneg1e1 12389	-1 · -1 is 1. (Contributed by David A. Wheeler, 8-Dec-2018.)
⊢ (-1 · -1) = 1
Theorem	halfre 12390	One-half is real. (Contributed by David A. Wheeler, 8-Dec-2018.)
⊢ (1 / 2) ∈ ℝ
Theorem	halfcn 12391	One-half is a complex number. (Contributed by David A. Wheeler, 8-Dec-2018.)
⊢ (1 / 2) ∈ ℂ
Theorem	halfgt0 12392	One-half is greater than zero. (Contributed by NM, 24-Feb-2005.)
⊢ 0 < (1 / 2)
Theorem	halfge0 12393	One-half is not negative. (Contributed by AV, 7-Jun-2020.)
⊢ 0 ≤ (1 / 2)
Theorem	halflt1 12394	One-half is less than one. (Contributed by NM, 24-Feb-2005.)
⊢ (1 / 2) < 1
Theorem	2halves 12395	Two halves make a whole. (Contributed by NM, 11-Apr-2005.)
⊢ (𝐴 ∈ ℂ → ((𝐴 / 2) + (𝐴 / 2)) = 𝐴)
Theorem	1mhlfehlf 12396	Prove that 1 - 1/2 = 1/2. (Contributed by David A. Wheeler, 4-Jan-2017.) (Proof shortened by SN, 22-Oct-2025.)
⊢ (1 − (1 / 2)) = (1 / 2)
Theorem	8th4div3 12397	An eighth of four thirds is a sixth. (Contributed by Paul Chapman, 24-Nov-2007.)
⊢ ((1 / 8) · (4 / 3)) = (1 / 6)
Theorem	halfthird 12398	Half minus a third. (Contributed by Scott Fenton, 8-Jul-2015.)
⊢ ((1 / 2) − (1 / 3)) = (1 / 6)
Theorem	halfpm6th 12399	One half plus or minus one sixth. (Contributed by Paul Chapman, 17-Jan-2008.) (Proof shortened by SN, 22-Oct-2025.)
⊢ (((1 / 2) − (1 / 6)) = (1 / 3) ∧ ((1 / 2) + (1 / 6)) = (2 / 3))
Theorem	it0e0 12400	i times 0 equals 0. (Contributed by David A. Wheeler, 8-Dec-2018.)
⊢ (i · 0) = 0