1p2e3 - Metamath Proof Explorer

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Theorem 1p2e3 12409

Description: 1 + 2 = 3. For a shorter proof using addcomli 11453, see 1p2e3ALT 12410. (Contributed by David A. Wheeler, 8-Dec-2018.) Reduce dependencies on axioms. (Revised by Steven Nguyen, 12-Dec-2022.)

**Assertion**
Ref	Expression
1p2e3	⊢ (1 + 2) = 3

**Proof of Theorem 1p2e3**
Step	Hyp	Ref	Expression
1		df-2 12329	. . 3 ⊢ 2 = (1 + 1)
2	1	oveq2i 7442	. 2 ⊢ (1 + 2) = (1 + (1 + 1))
3		ax-1cn 11213	. . 3 ⊢ 1 ∈ ℂ
4	3, 3, 3	addassi 11271	. 2 ⊢ ((1 + 1) + 1) = (1 + (1 + 1))
5		1p1e2 12391	. . . 4 ⊢ (1 + 1) = 2
6	5	oveq1i 7441	. . 3 ⊢ ((1 + 1) + 1) = (2 + 1)
7		2p1e3 12408	. . 3 ⊢ (2 + 1) = 3
8	6, 7	eqtri 2765	. 2 ⊢ ((1 + 1) + 1) = 3
9	2, 4, 8	3eqtr2i 2771	1 ⊢ (1 + 2) = 3

Colors of variables: wff setvar class

Syntax hints: = wceq 1540 (class class class)co 7431 1c1 11156 + caddc 11158 2c2 12321 3c3 12322

This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1795 ax-4 1809 ax-5 1910 ax-6 1967 ax-7 2007 ax-8 2110 ax-9 2118 ax-ext 2708 ax-1cn 11213 ax-addass 11220

This theorem depends on definitions: df-bi 207 df-an 396 df-or 849 df-3an 1089 df-tru 1543 df-fal 1553 df-ex 1780 df-sb 2065 df-clab 2715 df-cleq 2729 df-clel 2816 df-rab 3437 df-v 3482 df-dif 3954 df-un 3956 df-ss 3968 df-nul 4334 df-if 4526 df-sn 4627 df-pr 4629 df-op 4633 df-uni 4908 df-br 5144 df-iota 6514 df-fv 6569 df-ov 7434 df-2 12329 df-3 12330

This theorem is referenced by: fzo1to4tp 13793 binom3 14263 3lcm2e6woprm 16652 prmgaplem7 17095 2exp16 17128 prmlem1a 17144 23prm 17156 prmlem2 17157 83prm 17160 139prm 17161 163prm 17162 317prm 17163 631prm 17164 1259lem4 17171 1259prm 17173 2503lem2 17175 2503lem3 17176 4001lem2 17179 quart1lem 26898 log2ublem3 26991 log2ub 26992 pntibndlem2 27635 1kp2ke3k 30465 ex-ind-dvds 30480 fib4 34406 2np3bcnp1 42145 2xp3dxp2ge1d 42242 ex-decpmul 42340 sn-0ne2 42436 3cubeslem3r 42698 rabren3dioph 42826 fmtno4nprmfac193 47561 139prmALT 47583 127prm 47586 nnsum4primesodd 47783 nnsum4primesoddALTV 47784 ackval1012 48611

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