| 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 12398 3p2e5 12391 eqtri 2792 oveq1i 7421. (Contributed by Filip Cernatescu, 16-Mar-2019.) (Proof modification is discouraged.) |
| Ref | Expression |
|---|---|
| problem1 | ⊢ ((3 + 2) + 4) = 9 |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | 3p2e5 12391 | . . 3 ⊢ (3 + 2) = 5 | |
| 2 | 1 | oveq1i 7421 | . 2 ⊢ ((3 + 2) + 4) = (5 + 4) |
| 3 | 5p4e9 12398 | . 2 ⊢ (5 + 4) = 9 | |
| 4 | 2, 3 | eqtri 2792 | 1 ⊢ ((3 + 2) + 4) = 9 |
| Colors of variables: wff setvar class |
| Syntax hints: = wceq 1567 (class class class)co 7411 + caddc 11103 2c2 12295 3c3 12296 4c4 12297 5c5 12298 9c9 12302 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1822 ax-4 1836 ax-5 1937 ax-6 1994 ax-7 2035 ax-8 2151 ax-9 2159 ax-ext 2741 ax-1cn 11158 ax-addcl 11160 ax-addass 11165 |
| This theorem depends on definitions: df-bi 210 df-an 401 df-or 861 df-3an 1103 df-tru 1570 df-fal 1580 df-ex 1807 df-sb 2098 df-clab 2748 df-cleq 2761 df-clel 2844 df-rab 3424 df-v 3465 df-dif 3916 df-un 3918 df-ss 3930 df-nul 4295 df-if 4493 df-sn 4595 df-pr 4597 df-op 4601 df-uni 4877 df-br 5114 df-iota 6493 df-fv 6545 df-ov 7414 df-2 12303 df-3 12304 df-4 12305 df-5 12306 df-6 12307 df-7 12308 df-8 12309 df-9 12310 |
| This theorem is referenced by: (None) |
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