| 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 12334 3p2e5 12327 eqtri 2759 oveq1i 7377. (Contributed by Filip Cernatescu, 16-Mar-2019.) (Proof modification is discouraged.) |
| Ref | Expression |
|---|---|
| problem1 | ⊢ ((3 + 2) + 4) = 9 |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | 3p2e5 12327 | . . 3 ⊢ (3 + 2) = 5 | |
| 2 | 1 | oveq1i 7377 | . 2 ⊢ ((3 + 2) + 4) = (5 + 4) |
| 3 | 5p4e9 12334 | . 2 ⊢ (5 + 4) = 9 | |
| 4 | 2, 3 | eqtri 2759 | 1 ⊢ ((3 + 2) + 4) = 9 |
| Colors of variables: wff setvar class |
| Syntax hints: = wceq 1542 (class class class)co 7367 + caddc 11041 2c2 12236 3c3 12237 4c4 12238 5c5 12239 9c9 12243 |
| 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 2708 ax-1cn 11096 ax-addcl 11098 ax-addass 11103 |
| 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 2715 df-cleq 2728 df-clel 2811 df-rab 3390 df-v 3431 df-dif 3892 df-un 3894 df-ss 3906 df-nul 4274 df-if 4467 df-sn 4568 df-pr 4570 df-op 4574 df-uni 4851 df-br 5086 df-iota 6454 df-fv 6506 df-ov 7370 df-2 12244 df-3 12245 df-4 12246 df-5 12247 df-6 12248 df-7 12249 df-8 12250 df-9 12251 |
| This theorem is referenced by: (None) |
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