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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 12422 3p2e5 12415 eqtri 2763 oveq1i 7441. (Contributed by Filip Cernatescu, 16-Mar-2019.) (Proof modification is discouraged.) |
Ref | Expression |
---|---|
problem1 | ⊢ ((3 + 2) + 4) = 9 |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | 3p2e5 12415 | . . 3 ⊢ (3 + 2) = 5 | |
2 | 1 | oveq1i 7441 | . 2 ⊢ ((3 + 2) + 4) = (5 + 4) |
3 | 5p4e9 12422 | . 2 ⊢ (5 + 4) = 9 | |
4 | 2, 3 | eqtri 2763 | 1 ⊢ ((3 + 2) + 4) = 9 |
Colors of variables: wff setvar class |
Syntax hints: = wceq 1537 (class class class)co 7431 + caddc 11156 2c2 12319 3c3 12320 4c4 12321 5c5 12322 9c9 12326 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1792 ax-4 1806 ax-5 1908 ax-6 1965 ax-7 2005 ax-8 2108 ax-9 2116 ax-ext 2706 ax-1cn 11211 ax-addcl 11213 ax-addass 11218 |
This theorem depends on definitions: df-bi 207 df-an 396 df-or 848 df-3an 1088 df-tru 1540 df-fal 1550 df-ex 1777 df-sb 2063 df-clab 2713 df-cleq 2727 df-clel 2814 df-rab 3434 df-v 3480 df-dif 3966 df-un 3968 df-ss 3980 df-nul 4340 df-if 4532 df-sn 4632 df-pr 4634 df-op 4638 df-uni 4913 df-br 5149 df-iota 6516 df-fv 6571 df-ov 7434 df-2 12327 df-3 12328 df-4 12329 df-5 12330 df-6 12331 df-7 12332 df-8 12333 df-9 12334 |
This theorem is referenced by: (None) |
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