| 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 12403 3p2e5 12396 eqtri 2759 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 12396 | . . 3 ⊢ (3 + 2) = 5 | |
| 2 | 1 | oveq1i 7420 | . 2 ⊢ ((3 + 2) + 4) = (5 + 4) |
| 3 | 5p4e9 12403 | . 2 ⊢ (5 + 4) = 9 | |
| 4 | 2, 3 | eqtri 2759 | 1 ⊢ ((3 + 2) + 4) = 9 |
| Colors of variables: wff setvar class |
| Syntax hints: = wceq 1540 (class class class)co 7410 + caddc 11137 2c2 12300 3c3 12301 4c4 12302 5c5 12303 9c9 12307 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1795 ax-4 1809 ax-5 1910 ax-6 1967 ax-7 2008 ax-8 2111 ax-9 2119 ax-ext 2708 ax-1cn 11192 ax-addcl 11194 ax-addass 11199 |
| This theorem depends on definitions: df-bi 207 df-an 396 df-or 848 df-3an 1088 df-tru 1543 df-fal 1553 df-ex 1780 df-sb 2066 df-clab 2715 df-cleq 2728 df-clel 2810 df-rab 3421 df-v 3466 df-dif 3934 df-un 3936 df-ss 3948 df-nul 4314 df-if 4506 df-sn 4607 df-pr 4609 df-op 4613 df-uni 4889 df-br 5125 df-iota 6489 df-fv 6544 df-ov 7413 df-2 12308 df-3 12309 df-4 12310 df-5 12311 df-6 12312 df-7 12313 df-8 12314 df-9 12315 |
| This theorem is referenced by: (None) |
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