| 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 12278 3p2e5 12271 eqtri 2754 oveq1i 7356. (Contributed by Filip Cernatescu, 16-Mar-2019.) (Proof modification is discouraged.) |
| Ref | Expression |
|---|---|
| problem1 | ⊢ ((3 + 2) + 4) = 9 |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | 3p2e5 12271 | . . 3 ⊢ (3 + 2) = 5 | |
| 2 | 1 | oveq1i 7356 | . 2 ⊢ ((3 + 2) + 4) = (5 + 4) |
| 3 | 5p4e9 12278 | . 2 ⊢ (5 + 4) = 9 | |
| 4 | 2, 3 | eqtri 2754 | 1 ⊢ ((3 + 2) + 4) = 9 |
| Colors of variables: wff setvar class |
| Syntax hints: = wceq 1541 (class class class)co 7346 + caddc 11009 2c2 12180 3c3 12181 4c4 12182 5c5 12183 9c9 12187 |
| 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 2113 ax-9 2121 ax-ext 2703 ax-1cn 11064 ax-addcl 11066 ax-addass 11071 |
| 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 2710 df-cleq 2723 df-clel 2806 df-rab 3396 df-v 3438 df-dif 3900 df-un 3902 df-ss 3914 df-nul 4281 df-if 4473 df-sn 4574 df-pr 4576 df-op 4580 df-uni 4857 df-br 5090 df-iota 6437 df-fv 6489 df-ov 7349 df-2 12188 df-3 12189 df-4 12190 df-5 12191 df-6 12192 df-7 12193 df-8 12194 df-9 12195 |
| This theorem is referenced by: (None) |
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