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Mirrors > Home > MPE Home > Th. List > Mathboxes > problem1 | Structured version Visualization version GIF version |
Description: Practice problem 1. Clues: 5p4e9 11796 3p2e5 11789 eqtri 2844 oveq1i 7166. (Contributed by Filip Cernatescu, 16-Mar-2019.) (Proof modification is discouraged.) |
Ref | Expression |
---|---|
problem1 | ⊢ ((3 + 2) + 4) = 9 |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | 3p2e5 11789 | . . 3 ⊢ (3 + 2) = 5 | |
2 | 1 | oveq1i 7166 | . 2 ⊢ ((3 + 2) + 4) = (5 + 4) |
3 | 5p4e9 11796 | . 2 ⊢ (5 + 4) = 9 | |
4 | 2, 3 | eqtri 2844 | 1 ⊢ ((3 + 2) + 4) = 9 |
Colors of variables: wff setvar class |
Syntax hints: = wceq 1537 (class class class)co 7156 + caddc 10540 2c2 11693 3c3 11694 4c4 11695 5c5 11696 9c9 11700 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1796 ax-4 1810 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2116 ax-9 2124 ax-10 2145 ax-11 2161 ax-12 2177 ax-ext 2793 ax-1cn 10595 ax-addcl 10597 ax-addass 10602 |
This theorem depends on definitions: df-bi 209 df-an 399 df-or 844 df-3an 1085 df-tru 1540 df-ex 1781 df-nf 1785 df-sb 2070 df-clab 2800 df-cleq 2814 df-clel 2893 df-nfc 2963 df-rab 3147 df-v 3496 df-dif 3939 df-un 3941 df-in 3943 df-ss 3952 df-nul 4292 df-if 4468 df-sn 4568 df-pr 4570 df-op 4574 df-uni 4839 df-br 5067 df-iota 6314 df-fv 6363 df-ov 7159 df-2 11701 df-3 11702 df-4 11703 df-5 11704 df-6 11705 df-7 11706 df-8 11707 df-9 11708 |
This theorem is referenced by: (None) |
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