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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 2753 oveq1i 7429. (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 7429 | . 2 ⊢ ((3 + 2) + 4) = (5 + 4) |
3 | 5p4e9 12403 | . 2 ⊢ (5 + 4) = 9 | |
4 | 2, 3 | eqtri 2753 | 1 ⊢ ((3 + 2) + 4) = 9 |
Colors of variables: wff setvar class |
Syntax hints: = wceq 1533 (class class class)co 7419 + caddc 11143 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 1789 ax-4 1803 ax-5 1905 ax-6 1963 ax-7 2003 ax-8 2100 ax-9 2108 ax-ext 2696 ax-1cn 11198 ax-addcl 11200 ax-addass 11205 |
This theorem depends on definitions: df-bi 206 df-an 395 df-or 846 df-3an 1086 df-tru 1536 df-fal 1546 df-ex 1774 df-sb 2060 df-clab 2703 df-cleq 2717 df-clel 2802 df-rab 3419 df-v 3463 df-dif 3947 df-un 3949 df-ss 3961 df-nul 4323 df-if 4531 df-sn 4631 df-pr 4633 df-op 4637 df-uni 4910 df-br 5150 df-iota 6501 df-fv 6557 df-ov 7422 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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