| 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 12325 3p2e5 12318 eqtri 2762 oveq1i 7366. (Contributed by Filip Cernatescu, 16-Mar-2019.) (Proof modification is discouraged.) |
| Ref | Expression |
|---|---|
| problem1 | ⊢ ((3 + 2) + 4) = 9 |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | 3p2e5 12318 | . . 3 ⊢ (3 + 2) = 5 | |
| 2 | 1 | oveq1i 7366 | . 2 ⊢ ((3 + 2) + 4) = (5 + 4) |
| 3 | 5p4e9 12325 | . 2 ⊢ (5 + 4) = 9 | |
| 4 | 2, 3 | eqtri 2762 | 1 ⊢ ((3 + 2) + 4) = 9 |
| Colors of variables: wff setvar class |
| Syntax hints: = wceq 1547 (class class class)co 7356 + caddc 11032 2c2 12227 3c3 12228 4c4 12229 5c5 12230 9c9 12234 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1802 ax-4 1816 ax-5 1917 ax-6 1974 ax-7 2015 ax-8 2121 ax-9 2129 ax-ext 2711 ax-1cn 11087 ax-addcl 11089 ax-addass 11094 |
| This theorem depends on definitions: df-bi 208 df-an 397 df-or 854 df-3an 1094 df-tru 1550 df-fal 1560 df-ex 1787 df-sb 2074 df-clab 2718 df-cleq 2731 df-clel 2814 df-rab 3392 df-v 3433 df-dif 3886 df-un 3888 df-ss 3900 df-nul 4262 df-if 4455 df-sn 4556 df-pr 4558 df-op 4562 df-uni 4839 df-br 5073 df-iota 6441 df-fv 6493 df-ov 7359 df-2 12235 df-3 12236 df-4 12237 df-5 12238 df-6 12239 df-7 12240 df-8 12241 df-9 12242 |
| This theorem is referenced by: (None) |
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