| Mathbox for Filip Cernatescu |
< Previous
Next >
Nearby theorems |
||
| Mirrors > Home > MPE Home > Th. List > Mathboxes > problem1 | Structured version Visualization version GIF version | ||
| Description: Practice problem 1. Clues: 5p4e9 12328 3p2e5 12321 eqtri 2760 oveq1i 7371. (Contributed by Filip Cernatescu, 16-Mar-2019.) (Proof modification is discouraged.) |
| Ref | Expression |
|---|---|
| problem1 | ⊢ ((3 + 2) + 4) = 9 |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | 3p2e5 12321 | . . 3 ⊢ (3 + 2) = 5 | |
| 2 | 1 | oveq1i 7371 | . 2 ⊢ ((3 + 2) + 4) = (5 + 4) |
| 3 | 5p4e9 12328 | . 2 ⊢ (5 + 4) = 9 | |
| 4 | 2, 3 | eqtri 2760 | 1 ⊢ ((3 + 2) + 4) = 9 |
| Colors of variables: wff setvar class |
| Syntax hints: = wceq 1542 (class class class)co 7361 + caddc 11035 2c2 12230 3c3 12231 4c4 12232 5c5 12233 9c9 12237 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1912 ax-6 1969 ax-7 2010 ax-8 2116 ax-9 2124 ax-ext 2709 ax-1cn 11090 ax-addcl 11092 ax-addass 11097 |
| This theorem depends on definitions: df-bi 207 df-an 396 df-or 849 df-3an 1089 df-tru 1545 df-fal 1555 df-ex 1782 df-sb 2069 df-clab 2716 df-cleq 2729 df-clel 2812 df-rab 3391 df-v 3432 df-dif 3893 df-un 3895 df-ss 3907 df-nul 4275 df-if 4468 df-sn 4569 df-pr 4571 df-op 4575 df-uni 4852 df-br 5087 df-iota 6449 df-fv 6501 df-ov 7364 df-2 12238 df-3 12239 df-4 12240 df-5 12241 df-6 12242 df-7 12243 df-8 12244 df-9 12245 |
| This theorem is referenced by: (None) |
| Copyright terms: Public domain | W3C validator |