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 11044 3p2e5 11037 eqtri 2632 oveq1i 6559. (Contributed by Filip Cernatescu, 16-Mar-2019.) (Proof modification is discouraged.) |
Ref | Expression |
---|---|
problem1 | ⊢ ((3 + 2) + 4) = 9 |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | 3p2e5 11037 | . . 3 ⊢ (3 + 2) = 5 | |
2 | 1 | oveq1i 6559 | . 2 ⊢ ((3 + 2) + 4) = (5 + 4) |
3 | 5p4e9 11044 | . 2 ⊢ (5 + 4) = 9 | |
4 | 2, 3 | eqtri 2632 | 1 ⊢ ((3 + 2) + 4) = 9 |
Colors of variables: wff setvar class |
Syntax hints: = wceq 1475 (class class class)co 6549 + caddc 9818 2c2 10947 3c3 10948 4c4 10949 5c5 10950 9c9 10954 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-resscn 9872 ax-1cn 9873 ax-icn 9874 ax-addcl 9875 ax-addrcl 9876 ax-mulcl 9877 ax-mulrcl 9878 ax-addass 9880 ax-i2m1 9883 ax-1ne0 9884 ax-rrecex 9887 ax-cnre 9888 |
This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3an 1033 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-ral 2901 df-rex 2902 df-rab 2905 df-v 3175 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-nul 3875 df-if 4037 df-sn 4126 df-pr 4128 df-op 4132 df-uni 4373 df-br 4584 df-iota 5768 df-fv 5812 df-ov 6552 df-2 10956 df-3 10957 df-4 10958 df-5 10959 df-6 10960 df-7 10961 df-8 10962 df-9 10963 |
This theorem is referenced by: (None) |
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