| 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 12315 3p2e5 12308 eqtri 2752 oveq1i 7379. (Contributed by Filip Cernatescu, 16-Mar-2019.) (Proof modification is discouraged.) |
| Ref | Expression |
|---|---|
| problem1 | ⊢ ((3 + 2) + 4) = 9 |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | 3p2e5 12308 | . . 3 ⊢ (3 + 2) = 5 | |
| 2 | 1 | oveq1i 7379 | . 2 ⊢ ((3 + 2) + 4) = (5 + 4) |
| 3 | 5p4e9 12315 | . 2 ⊢ (5 + 4) = 9 | |
| 4 | 2, 3 | eqtri 2752 | 1 ⊢ ((3 + 2) + 4) = 9 |
| Colors of variables: wff setvar class |
| Syntax hints: = wceq 1540 (class class class)co 7369 + caddc 11047 2c2 12217 3c3 12218 4c4 12219 5c5 12220 9c9 12224 |
| 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 2701 ax-1cn 11102 ax-addcl 11104 ax-addass 11109 |
| 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 2708 df-cleq 2721 df-clel 2803 df-rab 3403 df-v 3446 df-dif 3914 df-un 3916 df-ss 3928 df-nul 4293 df-if 4485 df-sn 4586 df-pr 4588 df-op 4592 df-uni 4868 df-br 5103 df-iota 6452 df-fv 6507 df-ov 7372 df-2 12225 df-3 12226 df-4 12227 df-5 12228 df-6 12229 df-7 12230 df-8 12231 df-9 12232 |
| This theorem is referenced by: (None) |
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