| 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 12393 3p2e5 12386 eqtri 2786 oveq1i 7420. (Contributed by Filip Cernatescu, 16-Mar-2019.) (Proof modification is discouraged.) |
| Ref | Expression |
|---|---|
| problem1 | ⊢ ((3 + 2) + 4) = 9 |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | 3p2e5 12386 | . . 3 ⊢ (3 + 2) = 5 | |
| 2 | 1 | oveq1i 7420 | . 2 ⊢ ((3 + 2) + 4) = (5 + 4) |
| 3 | 5p4e9 12393 | . 2 ⊢ (5 + 4) = 9 | |
| 4 | 2, 3 | eqtri 2786 | 1 ⊢ ((3 + 2) + 4) = 9 |
| Colors of variables: wff setvar class |
| Syntax hints: = wceq 1570 (class class class)co 7410 + caddc 11098 2c2 12290 3c3 12291 4c4 12292 5c5 12293 9c9 12297 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1825 ax-4 1839 ax-5 1940 ax-6 1997 ax-7 2038 ax-8 2145 ax-9 2153 ax-ext 2735 ax-1cn 11153 ax-addcl 11155 ax-addass 11160 |
| This theorem depends on definitions: df-bi 210 df-an 401 df-or 861 df-3an 1105 df-tru 1573 df-fal 1583 df-ex 1810 df-sb 2097 df-clab 2742 df-cleq 2755 df-clel 2838 df-rab 3417 df-v 3457 df-dif 3908 df-un 3910 df-ss 3922 df-nul 4287 df-if 4488 df-sn 4590 df-pr 4592 df-op 4596 df-uni 4873 df-br 5110 df-iota 6492 df-fv 6544 df-ov 7413 df-2 12298 df-3 12299 df-4 12300 df-5 12301 df-6 12302 df-7 12303 df-8 12304 df-9 12305 |
| This theorem is referenced by: (None) |
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