| 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 12300 3p2e5 12293 eqtri 2759 oveq1i 7368. (Contributed by Filip Cernatescu, 16-Mar-2019.) (Proof modification is discouraged.) |
| Ref | Expression |
|---|---|
| problem1 | ⊢ ((3 + 2) + 4) = 9 |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | 3p2e5 12293 | . . 3 ⊢ (3 + 2) = 5 | |
| 2 | 1 | oveq1i 7368 | . 2 ⊢ ((3 + 2) + 4) = (5 + 4) |
| 3 | 5p4e9 12300 | . 2 ⊢ (5 + 4) = 9 | |
| 4 | 2, 3 | eqtri 2759 | 1 ⊢ ((3 + 2) + 4) = 9 |
| Colors of variables: wff setvar class |
| Syntax hints: = wceq 1541 (class class class)co 7358 + caddc 11031 2c2 12202 3c3 12203 4c4 12204 5c5 12205 9c9 12209 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1796 ax-4 1810 ax-5 1911 ax-6 1968 ax-7 2009 ax-8 2115 ax-9 2123 ax-ext 2708 ax-1cn 11086 ax-addcl 11088 ax-addass 11093 |
| This theorem depends on definitions: df-bi 207 df-an 396 df-or 848 df-3an 1088 df-tru 1544 df-fal 1554 df-ex 1781 df-sb 2068 df-clab 2715 df-cleq 2728 df-clel 2811 df-rab 3400 df-v 3442 df-dif 3904 df-un 3906 df-ss 3918 df-nul 4286 df-if 4480 df-sn 4581 df-pr 4583 df-op 4587 df-uni 4864 df-br 5099 df-iota 6448 df-fv 6500 df-ov 7361 df-2 12210 df-3 12211 df-4 12212 df-5 12213 df-6 12214 df-7 12215 df-8 12216 df-9 12217 |
| This theorem is referenced by: (None) |
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