Metamath Proof Explorer


Theorem line2ylem

Description: Lemma for line2y . This proof is based on counterexamples for the following cases: 1. C =/= 0 : p = (0,0) (LHS of biconditional is false, RHS is true); 2. C = 0 /\ B =/= 0 : p = (1,-A/B) (LHS of biconditional is true, RHS is false); 3. A = B = C = 0 : p = (1,1) (LHS of biconditional is true, RHS is false). (Contributed by AV, 4-Feb-2023)

Ref Expression
Hypotheses line2ylem.i ⊢ I = 1 2
line2ylem.p ⊢ P = ℝ I
Assertion line2ylem ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∀ p ∈ P A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0 → A ≠ 0 ∧ B = 0 ∧ C = 0

Proof

Step Hyp Ref Expression
1 line2ylem.i ⊢ I = 1 2
2 line2ylem.p ⊢ P = ℝ I
3 ianor ⊢ ¬ A ≠ 0 ∧ B = 0 ∧ C = 0 ↔ ¬ A ≠ 0 ∧ B = 0 ∨ ¬ C = 0
4 df-ne ⊢ C ≠ 0 ↔ ¬ C = 0
5 0re ⊢ 0 ∈ ℝ
6 1 2 prelrrx2 ⊢ 0 ∈ ℝ ∧ 0 ∈ ℝ → 1 0 2 0 ∈ P
7 5 5 6 mp2an ⊢ 1 0 2 0 ∈ P
8 eqneqall ⊢ C = 0 → C ≠ 0 → ¬ 0 = 0
9 8 com12 ⊢ C ≠ 0 → C = 0 → ¬ 0 = 0
10 eqid ⊢ 0 = 0
11 10 pm2.24i ⊢ ¬ 0 = 0 → C = 0
12 9 11 impbid1 ⊢ C ≠ 0 → C = 0 ↔ ¬ 0 = 0
13 12 adantl ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ C ≠ 0 → C = 0 ↔ ¬ 0 = 0
14 xor3 ⊢ ¬ C = 0 ↔ 0 = 0 ↔ C = 0 ↔ ¬ 0 = 0
15 13 14 sylibr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ C ≠ 0 → ¬ C = 0 ↔ 0 = 0
16 simp1 ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → A ∈ ℝ
17 16 recnd ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → A ∈ ℂ
18 17 mul01d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → A ⋅ 0 = 0
19 simp2 ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → B ∈ ℝ
20 19 recnd ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → B ∈ ℂ
21 20 mul01d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → B ⋅ 0 = 0
22 18 21 oveq12d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → A ⋅ 0 + B ⋅ 0 = 0 + 0
23 00id ⊢ 0 + 0 = 0
24 22 23 eqtrdi ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → A ⋅ 0 + B ⋅ 0 = 0
25 24 eqeq1d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → A ⋅ 0 + B ⋅ 0 = C ↔ 0 = C
26 eqcom ⊢ 0 = C ↔ C = 0
27 25 26 bitrdi ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → A ⋅ 0 + B ⋅ 0 = C ↔ C = 0
28 27 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ C ≠ 0 → A ⋅ 0 + B ⋅ 0 = C ↔ C = 0
29 28 bibi1d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ C ≠ 0 → A ⋅ 0 + B ⋅ 0 = C ↔ 0 = 0 ↔ C = 0 ↔ 0 = 0
30 15 29 mtbird ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ C ≠ 0 → ¬ A ⋅ 0 + B ⋅ 0 = C ↔ 0 = 0
31 fveq1 ⊢ p = 1 0 2 0 → p ⁡ 1 = 1 0 2 0 ⁡ 1
32 1ex ⊢ 1 ∈ V
33 c0ex ⊢ 0 ∈ V
34 1ne2 ⊢ 1 ≠ 2
35 fvpr1g ⊢ 1 ∈ V ∧ 0 ∈ V ∧ 1 ≠ 2 → 1 0 2 0 ⁡ 1 = 0
36 32 33 34 35 mp3an ⊢ 1 0 2 0 ⁡ 1 = 0
37 31 36 eqtrdi ⊢ p = 1 0 2 0 → p ⁡ 1 = 0
38 37 oveq2d ⊢ p = 1 0 2 0 → A ⁢ p ⁡ 1 = A ⋅ 0
39 fveq1 ⊢ p = 1 0 2 0 → p ⁡ 2 = 1 0 2 0 ⁡ 2
40 2ex ⊢ 2 ∈ V
41 fvpr2g ⊢ 2 ∈ V ∧ 0 ∈ V ∧ 1 ≠ 2 → 1 0 2 0 ⁡ 2 = 0
42 40 33 34 41 mp3an ⊢ 1 0 2 0 ⁡ 2 = 0
43 39 42 eqtrdi ⊢ p = 1 0 2 0 → p ⁡ 2 = 0
44 43 oveq2d ⊢ p = 1 0 2 0 → B ⁢ p ⁡ 2 = B ⋅ 0
45 38 44 oveq12d ⊢ p = 1 0 2 0 → A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = A ⋅ 0 + B ⋅ 0
46 45 eqeq1d ⊢ p = 1 0 2 0 → A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ A ⋅ 0 + B ⋅ 0 = C
47 37 eqeq1d ⊢ p = 1 0 2 0 → p ⁡ 1 = 0 ↔ 0 = 0
48 46 47 bibi12d ⊢ p = 1 0 2 0 → A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0 ↔ A ⋅ 0 + B ⋅ 0 = C ↔ 0 = 0
49 48 notbid ⊢ p = 1 0 2 0 → ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0 ↔ ¬ A ⋅ 0 + B ⋅ 0 = C ↔ 0 = 0
50 49 rspcev ⊢ 1 0 2 0 ∈ P ∧ ¬ A ⋅ 0 + B ⋅ 0 = C ↔ 0 = 0 → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
51 7 30 50 sylancr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ C ≠ 0 → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
52 51 expcom ⊢ C ≠ 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
53 4 52 sylbir ⊢ ¬ C = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
54 notnotb ⊢ C = 0 ↔ ¬ ¬ C = 0
55 ianor ⊢ ¬ A ≠ 0 ∧ B = 0 ↔ ¬ A ≠ 0 ∨ ¬ B = 0
56 df-ne ⊢ B ≠ 0 ↔ ¬ B = 0
57 1red ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → 1 ∈ ℝ
58 16 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → A ∈ ℝ
59 58 renegcld ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → − A ∈ ℝ
60 19 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → B ∈ ℝ
61 simprl ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → B ≠ 0
62 59 60 61 redivcld ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → − A B ∈ ℝ
63 1 2 prelrrx2 ⊢ 1 ∈ ℝ ∧ − A B ∈ ℝ → 1 1 2 − A B ∈ P
64 57 62 63 syl2anc ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → 1 1 2 − A B ∈ P
65 ax-1ne0 ⊢ 1 ≠ 0
66 65 neii ⊢ ¬ 1 = 0
67 10 66 2th ⊢ 0 = 0 ↔ ¬ 1 = 0
68 xor3 ⊢ ¬ 0 = 0 ↔ 1 = 0 ↔ 0 = 0 ↔ ¬ 1 = 0
69 67 68 mpbir ⊢ ¬ 0 = 0 ↔ 1 = 0
70 17 mulridd ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → A ⋅ 1 = A
71 70 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → A ⋅ 1 = A
72 17 negcld ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → − A ∈ ℂ
73 72 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → − A ∈ ℂ
74 20 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → B ∈ ℂ
75 73 74 61 divcan2d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → B ⁢ − A B = − A
76 71 75 oveq12d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → A ⋅ 1 + B ⁢ − A B = A + − A
77 17 negidd ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → A + − A = 0
78 77 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → A + − A = 0
79 76 78 eqtrd ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → A ⋅ 1 + B ⁢ − A B = 0
80 simprr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → C = 0
81 79 80 eqeq12d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → A ⋅ 1 + B ⁢ − A B = C ↔ 0 = 0
82 81 bibi1d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → A ⋅ 1 + B ⁢ − A B = C ↔ 1 = 0 ↔ 0 = 0 ↔ 1 = 0
83 69 82 mtbiri ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → ¬ A ⋅ 1 + B ⁢ − A B = C ↔ 1 = 0
84 fveq1 ⊢ p = 1 1 2 − A B → p ⁡ 1 = 1 1 2 − A B ⁡ 1
85 fvpr1g ⊢ 1 ∈ V ∧ 1 ∈ V ∧ 1 ≠ 2 → 1 1 2 − A B ⁡ 1 = 1
86 32 32 34 85 mp3an ⊢ 1 1 2 − A B ⁡ 1 = 1
87 84 86 eqtrdi ⊢ p = 1 1 2 − A B → p ⁡ 1 = 1
88 87 oveq2d ⊢ p = 1 1 2 − A B → A ⁢ p ⁡ 1 = A ⋅ 1
89 fveq1 ⊢ p = 1 1 2 − A B → p ⁡ 2 = 1 1 2 − A B ⁡ 2
90 ovex ⊢ − A B ∈ V
91 fvpr2g ⊢ 2 ∈ V ∧ − A B ∈ V ∧ 1 ≠ 2 → 1 1 2 − A B ⁡ 2 = − A B
92 40 90 34 91 mp3an ⊢ 1 1 2 − A B ⁡ 2 = − A B
93 89 92 eqtrdi ⊢ p = 1 1 2 − A B → p ⁡ 2 = − A B
94 93 oveq2d ⊢ p = 1 1 2 − A B → B ⁢ p ⁡ 2 = B ⁢ − A B
95 88 94 oveq12d ⊢ p = 1 1 2 − A B → A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = A ⋅ 1 + B ⁢ − A B
96 95 eqeq1d ⊢ p = 1 1 2 − A B → A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ A ⋅ 1 + B ⁢ − A B = C
97 87 eqeq1d ⊢ p = 1 1 2 − A B → p ⁡ 1 = 0 ↔ 1 = 0
98 96 97 bibi12d ⊢ p = 1 1 2 − A B → A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0 ↔ A ⋅ 1 + B ⁢ − A B = C ↔ 1 = 0
99 98 notbid ⊢ p = 1 1 2 − A B → ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0 ↔ ¬ A ⋅ 1 + B ⁢ − A B = C ↔ 1 = 0
100 99 rspcev ⊢ 1 1 2 − A B ∈ P ∧ ¬ A ⋅ 1 + B ⁢ − A B = C ↔ 1 = 0 → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
101 64 83 100 syl2anc ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
102 101 expcom ⊢ B ≠ 0 ∧ C = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
103 102 ex ⊢ B ≠ 0 → C = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
104 56 103 sylbir ⊢ ¬ B = 0 → C = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
105 notnotb ⊢ B = 0 ↔ ¬ ¬ B = 0
106 nne ⊢ ¬ A ≠ 0 ↔ A = 0
107 106 bicomi ⊢ A = 0 ↔ ¬ A ≠ 0
108 1re ⊢ 1 ∈ ℝ
109 1 2 prelrrx2 ⊢ 1 ∈ ℝ ∧ 1 ∈ ℝ → 1 1 2 1 ∈ P
110 108 108 109 mp2an ⊢ 1 1 2 1 ∈ P
111 oveq1 ⊢ A = 0 → A ⋅ 1 = 0 ⋅ 1
112 111 adantl ⊢ B = 0 ∧ A = 0 → A ⋅ 1 = 0 ⋅ 1
113 ax-1cn ⊢ 1 ∈ ℂ
114 113 mul02i ⊢ 0 ⋅ 1 = 0
115 112 114 eqtrdi ⊢ B = 0 ∧ A = 0 → A ⋅ 1 = 0
116 oveq1 ⊢ B = 0 → B ⋅ 1 = 0 ⋅ 1
117 116 adantr ⊢ B = 0 ∧ A = 0 → B ⋅ 1 = 0 ⋅ 1
118 117 114 eqtrdi ⊢ B = 0 ∧ A = 0 → B ⋅ 1 = 0
119 115 118 oveq12d ⊢ B = 0 ∧ A = 0 → A ⋅ 1 + B ⋅ 1 = 0 + 0
120 119 23 eqtrdi ⊢ B = 0 ∧ A = 0 → A ⋅ 1 + B ⋅ 1 = 0
121 id ⊢ C = 0 → C = 0
122 120 121 eqeqan12d ⊢ B = 0 ∧ A = 0 ∧ C = 0 → A ⋅ 1 + B ⋅ 1 = C ↔ 0 = 0
123 122 bibi1d ⊢ B = 0 ∧ A = 0 ∧ C = 0 → A ⋅ 1 + B ⋅ 1 = C ↔ 1 = 0 ↔ 0 = 0 ↔ 1 = 0
124 69 123 mtbiri ⊢ B = 0 ∧ A = 0 ∧ C = 0 → ¬ A ⋅ 1 + B ⋅ 1 = C ↔ 1 = 0
125 fveq1 ⊢ p = 1 1 2 1 → p ⁡ 1 = 1 1 2 1 ⁡ 1
126 fvpr1g ⊢ 1 ∈ V ∧ 1 ∈ V ∧ 1 ≠ 2 → 1 1 2 1 ⁡ 1 = 1
127 32 32 34 126 mp3an ⊢ 1 1 2 1 ⁡ 1 = 1
128 125 127 eqtrdi ⊢ p = 1 1 2 1 → p ⁡ 1 = 1
129 128 oveq2d ⊢ p = 1 1 2 1 → A ⁢ p ⁡ 1 = A ⋅ 1
130 fveq1 ⊢ p = 1 1 2 1 → p ⁡ 2 = 1 1 2 1 ⁡ 2
131 fvpr2g ⊢ 2 ∈ V ∧ 1 ∈ V ∧ 1 ≠ 2 → 1 1 2 1 ⁡ 2 = 1
132 40 32 34 131 mp3an ⊢ 1 1 2 1 ⁡ 2 = 1
133 130 132 eqtrdi ⊢ p = 1 1 2 1 → p ⁡ 2 = 1
134 133 oveq2d ⊢ p = 1 1 2 1 → B ⁢ p ⁡ 2 = B ⋅ 1
135 129 134 oveq12d ⊢ p = 1 1 2 1 → A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = A ⋅ 1 + B ⋅ 1
136 135 eqeq1d ⊢ p = 1 1 2 1 → A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ A ⋅ 1 + B ⋅ 1 = C
137 128 eqeq1d ⊢ p = 1 1 2 1 → p ⁡ 1 = 0 ↔ 1 = 0
138 136 137 bibi12d ⊢ p = 1 1 2 1 → A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0 ↔ A ⋅ 1 + B ⋅ 1 = C ↔ 1 = 0
139 138 notbid ⊢ p = 1 1 2 1 → ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0 ↔ ¬ A ⋅ 1 + B ⋅ 1 = C ↔ 1 = 0
140 139 rspcev ⊢ 1 1 2 1 ∈ P ∧ ¬ A ⋅ 1 + B ⋅ 1 = C ↔ 1 = 0 → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
141 110 124 140 sylancr ⊢ B = 0 ∧ A = 0 ∧ C = 0 → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
142 141 a1d ⊢ B = 0 ∧ A = 0 ∧ C = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
143 142 ex ⊢ B = 0 ∧ A = 0 → C = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
144 105 107 143 syl2anbr ⊢ ¬ ¬ B = 0 ∧ ¬ A ≠ 0 → C = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
145 104 144 jaoi3 ⊢ ¬ B = 0 ∨ ¬ A ≠ 0 → C = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
146 145 orcoms ⊢ ¬ A ≠ 0 ∨ ¬ B = 0 → C = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
147 55 146 sylbi ⊢ ¬ A ≠ 0 ∧ B = 0 → C = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
148 147 com12 ⊢ C = 0 → ¬ A ≠ 0 ∧ B = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
149 54 148 sylbir ⊢ ¬ ¬ C = 0 → ¬ A ≠ 0 ∧ B = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
150 149 imp ⊢ ¬ ¬ C = 0 ∧ ¬ A ≠ 0 ∧ B = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
151 53 150 jaoi3 ⊢ ¬ C = 0 ∨ ¬ A ≠ 0 ∧ B = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
152 151 orcoms ⊢ ¬ A ≠ 0 ∧ B = 0 ∨ ¬ C = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
153 152 com12 ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ¬ A ≠ 0 ∧ B = 0 ∨ ¬ C = 0 → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
154 3 153 biimtrid ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ¬ A ≠ 0 ∧ B = 0 ∧ C = 0 → ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
155 rexnal ⊢ ∃ p ∈ P ¬ A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0 ↔ ¬ ∀ p ∈ P A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
156 154 155 imbitrdi ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ¬ A ≠ 0 ∧ B = 0 ∧ C = 0 → ¬ ∀ p ∈ P A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0
157 156 con4d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∀ p ∈ P A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0 → A ≠ 0 ∧ B = 0 ∧ C = 0
158 df-3an ⊢ A ≠ 0 ∧ B = 0 ∧ C = 0 ↔ A ≠ 0 ∧ B = 0 ∧ C = 0
159 157 158 imbitrrdi ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∀ p ∈ P A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ p ⁡ 1 = 0 → A ≠ 0 ∧ B = 0 ∧ C = 0