Users' Mathboxes Mathbox for Filip Cernatescu < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >   Mathboxes  >  problem1 Structured version   Unicode version

Theorem problem1 29248
Description: Practice problem 1. Clues: 5p4e9 10614 3p2e5 10607 eqtri 2425 oveq1i 6228. (Contributed by Filip Cernatescu, 16-Mar-2019.) (Proof modification is discouraged.)
Assertion
Ref Expression
problem1  |-  ( ( 3  +  2 )  +  4 )  =  9

Proof of Theorem problem1
StepHypRef Expression
1 3p2e5 10607 . . 3  |-  ( 3  +  2 )  =  5
21oveq1i 6228 . 2  |-  ( ( 3  +  2 )  +  4 )  =  ( 5  +  4 )
3 5p4e9 10614 . 2  |-  ( 5  +  4 )  =  9
42, 3eqtri 2425 1  |-  ( ( 3  +  2 )  +  4 )  =  9
Colors of variables: wff setvar class
Syntax hints:    = wceq 1399  (class class class)co 6218    + caddc 9428   2c2 10524   3c3 10525   4c4 10526   5c5 10527   9c9 10531
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1633  ax-4 1646  ax-5 1719  ax-6 1765  ax-7 1808  ax-10 1855  ax-11 1860  ax-12 1872  ax-13 2020  ax-ext 2374  ax-resscn 9482  ax-1cn 9483  ax-icn 9484  ax-addcl 9485  ax-addrcl 9486  ax-mulcl 9487  ax-mulrcl 9488  ax-addass 9490  ax-i2m1 9493  ax-1ne0 9494  ax-rrecex 9497  ax-cnre 9498
This theorem depends on definitions:  df-bi 185  df-or 368  df-an 369  df-3an 973  df-tru 1402  df-ex 1628  df-nf 1632  df-sb 1758  df-clab 2382  df-cleq 2388  df-clel 2391  df-nfc 2546  df-ne 2593  df-ral 2751  df-rex 2752  df-rab 2755  df-v 3053  df-dif 3409  df-un 3411  df-in 3413  df-ss 3420  df-nul 3729  df-if 3875  df-sn 3962  df-pr 3964  df-op 3968  df-uni 4181  df-br 4385  df-iota 5477  df-fv 5521  df-ov 6221  df-2 10533  df-3 10534  df-4 10535  df-5 10536  df-6 10537  df-7 10538  df-8 10539  df-9 10540
This theorem is referenced by: (None)
  Copyright terms: Public domain W3C validator