| Mathbox for Filip Cernatescu |
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| Mirrors > Home > MPE Home > Th. List > Mathboxes > problem1 | Structured version Visualization version GIF version | ||
| Description: Practice problem 1. Clues: 5p4e9 12376 3p2e5 12369 eqtri 2786 oveq1i 7407. (Contributed by Filip Cernatescu, 16-Mar-2019.) (Proof modification is discouraged.) |
| Ref | Expression |
|---|---|
| problem1 | ⊢ ((3 + 2) + 4) = 9 |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | 3p2e5 12369 | . . 3 ⊢ (3 + 2) = 5 | |
| 2 | 1 | oveq1i 7407 | . 2 ⊢ ((3 + 2) + 4) = (5 + 4) |
| 3 | 5p4e9 12376 | . 2 ⊢ (5 + 4) = 9 | |
| 4 | 2, 3 | eqtri 2786 | 1 ⊢ ((3 + 2) + 4) = 9 |
| Colors of variables: wff setvar class |
| Syntax hints: = wceq 1561 (class class class)co 7397 + caddc 11077 2c2 12273 3c3 12274 4c4 12275 5c5 12276 9c9 12280 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1816 ax-4 1830 ax-5 1931 ax-6 1988 ax-7 2029 ax-8 2145 ax-9 2153 ax-ext 2735 ax-1cn 11132 ax-addcl 11134 ax-addass 11139 |
| This theorem depends on definitions: df-bi 209 df-an 400 df-or 859 df-3an 1101 df-tru 1564 df-fal 1574 df-ex 1801 df-sb 2092 df-clab 2742 df-cleq 2755 df-clel 2838 df-rab 3416 df-v 3457 df-dif 3908 df-un 3910 df-ss 3922 df-nul 4287 df-if 4482 df-sn 4584 df-pr 4586 df-op 4590 df-uni 4867 df-br 5102 df-iota 6478 df-fv 6530 df-ov 7400 df-2 12281 df-3 12282 df-4 12283 df-5 12284 df-6 12285 df-7 12286 df-8 12287 df-9 12288 |
| This theorem is referenced by: (None) |
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