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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 11783 3p2e5 11776 eqtri 2821 oveq1i 7145. (Contributed by Filip Cernatescu, 16-Mar-2019.) (Proof modification is discouraged.) |
Ref | Expression |
---|---|
problem1 | ⊢ ((3 + 2) + 4) = 9 |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | 3p2e5 11776 | . . 3 ⊢ (3 + 2) = 5 | |
2 | 1 | oveq1i 7145 | . 2 ⊢ ((3 + 2) + 4) = (5 + 4) |
3 | 5p4e9 11783 | . 2 ⊢ (5 + 4) = 9 | |
4 | 2, 3 | eqtri 2821 | 1 ⊢ ((3 + 2) + 4) = 9 |
Colors of variables: wff setvar class |
Syntax hints: = wceq 1538 (class class class)co 7135 + caddc 10529 2c2 11680 3c3 11681 4c4 11682 5c5 11683 9c9 11687 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2113 ax-9 2121 ax-ext 2770 ax-1cn 10584 ax-addcl 10586 ax-addass 10591 |
This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3an 1086 df-ex 1782 df-sb 2070 df-clab 2777 df-cleq 2791 df-clel 2870 df-v 3443 df-un 3886 df-in 3888 df-ss 3898 df-sn 4526 df-pr 4528 df-op 4532 df-uni 4801 df-br 5031 df-iota 6283 df-fv 6332 df-ov 7138 df-2 11688 df-3 11689 df-4 11690 df-5 11691 df-6 11692 df-7 11693 df-8 11694 df-9 11695 |
This theorem is referenced by: (None) |
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