Metamath Proof Explorer


Theorem problem4

Description: Practice problem 4. Clues: pm3.2i eqcomi eqtri subaddrii recni 7re 6re ax-1cn df-7 ax-mp oveq1i 3cn 2cn df-3 mullidi subdiri mp3an mulcli subadd23 oveq2i oveq12i 3t2e6 mulcomi subcli biimpri subadd2i . (Contributed by Filip Cernatescu, 16-Mar-2019) (Proof modification is discouraged.)

Ref Expression
Hypotheses problem4.1 ⊢ A ∈ ℂ
problem4.2 ⊢ B ∈ ℂ
problem4.3 ⊢ A + B = 3
problem4.4 ⊢ 3 ⁢ A + 2 ⁢ B = 7
Assertion problem4 ⊢ A = 1 ∧ B = 2

Proof

Step Hyp Ref Expression
1 problem4.1 ⊢ A ∈ ℂ
2 problem4.2 ⊢ B ∈ ℂ
3 problem4.3 ⊢ A + B = 3
4 problem4.4 ⊢ 3 ⁢ A + 2 ⁢ B = 7
5 7re ⊢ 7 ∈ ℝ
6 5 recni ⊢ 7 ∈ ℂ
7 6re ⊢ 6 ∈ ℝ
8 7 recni ⊢ 6 ∈ ℂ
9 ax-1cn ⊢ 1 ∈ ℂ
10 df-7 ⊢ 7 = 6 + 1
11 10 eqcomi ⊢ 6 + 1 = 7
12 6 8 9 11 subaddrii ⊢ 7 − 6 = 1
13 12 eqcomi ⊢ 1 = 7 − 6
14 3cn ⊢ 3 ∈ ℂ
15 2cn ⊢ 2 ∈ ℂ
16 df-3 ⊢ 3 = 2 + 1
17 16 eqcomi ⊢ 2 + 1 = 3
18 14 15 9 17 subaddrii ⊢ 3 − 2 = 1
19 18 oveq1i ⊢ 3 − 2 ⁢ A = 1 ⁢ A
20 1 mullidi ⊢ 1 ⁢ A = A
21 19 20 eqtri ⊢ 3 − 2 ⁢ A = A
22 21 eqcomi ⊢ A = 3 − 2 ⁢ A
23 14 15 1 subdiri ⊢ 3 − 2 ⁢ A = 3 ⁢ A − 2 ⁢ A
24 22 23 eqtri ⊢ A = 3 ⁢ A − 2 ⁢ A
25 24 oveq1i ⊢ A + 6 = 3 ⁢ A - 2 ⁢ A + 6
26 14 1 mulcli ⊢ 3 ⁢ A ∈ ℂ
27 15 1 mulcli ⊢ 2 ⁢ A ∈ ℂ
28 subadd23 ⊢ 3 ⁢ A ∈ ℂ ∧ 2 ⁢ A ∈ ℂ ∧ 6 ∈ ℂ → 3 ⁢ A - 2 ⁢ A + 6 = 3 ⁢ A + 6 - 2 ⁢ A
29 26 27 8 28 mp3an ⊢ 3 ⁢ A - 2 ⁢ A + 6 = 3 ⁢ A + 6 - 2 ⁢ A
30 3t2e6 ⊢ 3 ⋅ 2 = 6
31 1 15 mulcomi ⊢ A ⋅ 2 = 2 ⁢ A
32 30 31 oveq12i ⊢ 3 ⋅ 2 − A ⋅ 2 = 6 − 2 ⁢ A
33 32 eqcomi ⊢ 6 − 2 ⁢ A = 3 ⋅ 2 − A ⋅ 2
34 14 1 15 subdiri ⊢ 3 − A ⋅ 2 = 3 ⋅ 2 − A ⋅ 2
35 34 eqcomi ⊢ 3 ⋅ 2 − A ⋅ 2 = 3 − A ⋅ 2
36 14 1 subcli ⊢ 3 − A ∈ ℂ
37 15 36 mulcomi ⊢ 2 ⁢ 3 − A = 3 − A ⋅ 2
38 37 eqcomi ⊢ 3 − A ⋅ 2 = 2 ⁢ 3 − A
39 14 1 2 3 subaddrii ⊢ 3 − A = B
40 39 eqcomi ⊢ B = 3 − A
41 40 oveq2i ⊢ 2 ⁢ B = 2 ⁢ 3 − A
42 41 eqcomi ⊢ 2 ⁢ 3 − A = 2 ⁢ B
43 38 42 eqtri ⊢ 3 − A ⋅ 2 = 2 ⁢ B
44 35 43 eqtri ⊢ 3 ⋅ 2 − A ⋅ 2 = 2 ⁢ B
45 33 44 eqtri ⊢ 6 − 2 ⁢ A = 2 ⁢ B
46 45 eqcomi ⊢ 2 ⁢ B = 6 − 2 ⁢ A
47 46 oveq2i ⊢ 3 ⁢ A + 2 ⁢ B = 3 ⁢ A + 6 - 2 ⁢ A
48 47 eqcomi ⊢ 3 ⁢ A + 6 - 2 ⁢ A = 3 ⁢ A + 2 ⁢ B
49 29 48 eqtri ⊢ 3 ⁢ A - 2 ⁢ A + 6 = 3 ⁢ A + 2 ⁢ B
50 25 49 eqtri ⊢ A + 6 = 3 ⁢ A + 2 ⁢ B
51 50 4 eqtri ⊢ A + 6 = 7
52 6 8 1 subadd2i ⊢ 7 − 6 = A ↔ A + 6 = 7
53 52 biimpri ⊢ A + 6 = 7 → 7 − 6 = A
54 51 53 ax-mp ⊢ 7 − 6 = A
55 13 54 eqtri ⊢ 1 = A
56 55 eqcomi ⊢ A = 1
57 56 oveq2i ⊢ 3 − A = 3 − 1
58 14 9 15 subadd2i ⊢ 3 − 1 = 2 ↔ 2 + 1 = 3
59 58 biimpri ⊢ 2 + 1 = 3 → 3 − 1 = 2
60 17 59 ax-mp ⊢ 3 − 1 = 2
61 57 60 eqtri ⊢ 3 − A = 2
62 40 61 eqtri ⊢ B = 2
63 56 62 pm3.2i ⊢ A = 1 ∧ B = 2