Theorem problem1 36196

Description: Practice problem 1. Clues: 5p4e9 12415 3p2e5 12408 eqtri 2788 oveq1i 7429. (Contributed by Filip Cernatescu, 16-Mar-2019.) (Proof modification is discouraged.)

Assertion

Ref	Expression
problem1	⊢ ((3 + 2) + 4) = 9

Proof of Theorem problem1

Step	Hyp	Ref	Expression
1		3p2e5 12408	. . 3 ⊢ (3 + 2) = 5
2	1	oveq1i 7429	. 2 ⊢ ((3 + 2) + 4) = (5 + 4)
3		5p4e9 12415	. 2 ⊢ (5 + 4) = 9
4	2, 3	eqtri 2788	1 ⊢ ((3 + 2) + 4) = 9

This proof depends on syntax axioms:   = wceq 1570  (class class class)co 7419   + caddc 11120  2c2 12312  3c3 12313  4c4 12314  5c5 12315  9c9 12319

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737  ax-1cn 11175  ax-addcl 11177  ax-addass 11182

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-ss 3923  df-nul 4287  df-if 4490  df-sn 4592  df-pr 4594  df-op 4598  df-uni 4875  df-br 5112  df-iota 6496  df-fv 6548  df-ov 7422  df-2 12320  df-3 12321  df-4 12322  df-5 12323  df-6 12324  df-7 12325  df-8 12326  df-9 12327

This theorem is used by: (None)