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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 11374 3p2e5 11367 eqtri 2793 oveq1i 6806. (Contributed by Filip Cernatescu, 16-Mar-2019.) (Proof modification is discouraged.) |
Ref | Expression |
---|---|
problem1 | ⊢ ((3 + 2) + 4) = 9 |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | 3p2e5 11367 | . . 3 ⊢ (3 + 2) = 5 | |
2 | 1 | oveq1i 6806 | . 2 ⊢ ((3 + 2) + 4) = (5 + 4) |
3 | 5p4e9 11374 | . 2 ⊢ (5 + 4) = 9 | |
4 | 2, 3 | eqtri 2793 | 1 ⊢ ((3 + 2) + 4) = 9 |
Colors of variables: wff setvar class |
Syntax hints: = wceq 1631 (class class class)co 6796 + caddc 10145 2c2 11276 3c3 11277 4c4 11278 5c5 11279 9c9 11283 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1870 ax-4 1885 ax-5 1991 ax-6 2057 ax-7 2093 ax-9 2154 ax-10 2174 ax-11 2190 ax-12 2203 ax-13 2408 ax-ext 2751 ax-resscn 10199 ax-1cn 10200 ax-icn 10201 ax-addcl 10202 ax-addrcl 10203 ax-mulcl 10204 ax-mulrcl 10205 ax-addass 10207 ax-i2m1 10210 ax-1ne0 10211 ax-rrecex 10214 ax-cnre 10215 |
This theorem depends on definitions: df-bi 197 df-an 383 df-or 837 df-3an 1073 df-tru 1634 df-ex 1853 df-nf 1858 df-sb 2050 df-clab 2758 df-cleq 2764 df-clel 2767 df-nfc 2902 df-ne 2944 df-ral 3066 df-rex 3067 df-rab 3070 df-v 3353 df-dif 3726 df-un 3728 df-in 3730 df-ss 3737 df-nul 4064 df-if 4227 df-sn 4318 df-pr 4320 df-op 4324 df-uni 4576 df-br 4788 df-iota 5993 df-fv 6038 df-ov 6799 df-2 11285 df-3 11286 df-4 11287 df-5 11288 df-6 11289 df-7 11290 df-8 11291 df-9 11292 |
This theorem is referenced by: (None) |
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