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

Proof

Step Hyp Ref Expression
1 line2ylem.i ⊢ I = 0 ..^ 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 → ⟨“ 0 0 ”⟩ ∈ P
7 5 5 6 mp2an ⊢ ⟨“ 0 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 = ⟨“ 0 0 ”⟩ → p ⁡ 0 = ⟨“ 0 0 ”⟩ ⁡ 0
32 c0ex ⊢ 0 ∈ V
33 s2fv0 ⊢ 0 ∈ V → ⟨“ 0 0 ”⟩ ⁡ 0 = 0
34 32 33 ax-mp ⊢ ⟨“ 0 0 ”⟩ ⁡ 0 = 0
35 31 34 eqtrdi ⊢ p = ⟨“ 0 0 ”⟩ → p ⁡ 0 = 0
36 35 oveq2d ⊢ p = ⟨“ 0 0 ”⟩ → A ⁢ p ⁡ 0 = A ⋅ 0
37 fveq1 ⊢ p = ⟨“ 0 0 ”⟩ → p ⁡ 1 = ⟨“ 0 0 ”⟩ ⁡ 1
38 s2fv1 ⊢ 0 ∈ V → ⟨“ 0 0 ”⟩ ⁡ 1 = 0
39 32 38 ax-mp ⊢ ⟨“ 0 0 ”⟩ ⁡ 1 = 0
40 37 39 eqtrdi ⊢ p = ⟨“ 0 0 ”⟩ → p ⁡ 1 = 0
41 40 oveq2d ⊢ p = ⟨“ 0 0 ”⟩ → B ⁢ p ⁡ 1 = B ⋅ 0
42 36 41 oveq12d ⊢ p = ⟨“ 0 0 ”⟩ → A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = A ⋅ 0 + B ⋅ 0
43 42 eqeq1d ⊢ p = ⟨“ 0 0 ”⟩ → A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ A ⋅ 0 + B ⋅ 0 = C
44 35 eqeq1d ⊢ p = ⟨“ 0 0 ”⟩ → p ⁡ 0 = 0 ↔ 0 = 0
45 43 44 bibi12d ⊢ p = ⟨“ 0 0 ”⟩ → A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0 ↔ A ⋅ 0 + B ⋅ 0 = C ↔ 0 = 0
46 45 notbid ⊢ p = ⟨“ 0 0 ”⟩ → ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0 ↔ ¬ A ⋅ 0 + B ⋅ 0 = C ↔ 0 = 0
47 46 rspcev ⊢ ⟨“ 0 0 ”⟩ ∈ P ∧ ¬ A ⋅ 0 + B ⋅ 0 = C ↔ 0 = 0 → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
48 7 30 47 sylancr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ C ≠ 0 → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
49 48 expcom ⊢ C ≠ 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
50 4 49 sylbir ⊢ ¬ C = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
51 notnotb ⊢ C = 0 ↔ ¬ ¬ C = 0
52 ianor ⊢ ¬ A ≠ 0 ∧ B = 0 ↔ ¬ A ≠ 0 ∨ ¬ B = 0
53 df-ne ⊢ B ≠ 0 ↔ ¬ B = 0
54 fveq1 ⊢ p = ⟨“ 1 − A B ”⟩ → p ⁡ 0 = ⟨“ 1 − A B ”⟩ ⁡ 0
55 1ex ⊢ 1 ∈ V
56 s2fv0 ⊢ 1 ∈ V → ⟨“ 1 − A B ”⟩ ⁡ 0 = 1
57 55 56 ax-mp ⊢ ⟨“ 1 − A B ”⟩ ⁡ 0 = 1
58 54 57 eqtrdi ⊢ p = ⟨“ 1 − A B ”⟩ → p ⁡ 0 = 1
59 58 oveq2d ⊢ p = ⟨“ 1 − A B ”⟩ → A ⁢ p ⁡ 0 = A ⋅ 1
60 fveq1 ⊢ p = ⟨“ 1 − A B ”⟩ → p ⁡ 1 = ⟨“ 1 − A B ”⟩ ⁡ 1
61 ovex ⊢ − A B ∈ V
62 s2fv1 ⊢ − A B ∈ V → ⟨“ 1 − A B ”⟩ ⁡ 1 = − A B
63 61 62 ax-mp ⊢ ⟨“ 1 − A B ”⟩ ⁡ 1 = − A B
64 60 63 eqtrdi ⊢ p = ⟨“ 1 − A B ”⟩ → p ⁡ 1 = − A B
65 64 oveq2d ⊢ p = ⟨“ 1 − A B ”⟩ → B ⁢ p ⁡ 1 = B ⁢ − A B
66 59 65 oveq12d ⊢ p = ⟨“ 1 − A B ”⟩ → A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = A ⋅ 1 + B ⁢ − A B
67 66 eqeq1d ⊢ p = ⟨“ 1 − A B ”⟩ → A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ A ⋅ 1 + B ⁢ − A B = C
68 58 eqeq1d ⊢ p = ⟨“ 1 − A B ”⟩ → p ⁡ 0 = 0 ↔ 1 = 0
69 67 68 bibi12d ⊢ p = ⟨“ 1 − A B ”⟩ → A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0 ↔ A ⋅ 1 + B ⁢ − A B = C ↔ 1 = 0
70 69 notbid ⊢ p = ⟨“ 1 − A B ”⟩ → ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0 ↔ ¬ A ⋅ 1 + B ⁢ − A B = C ↔ 1 = 0
71 1red ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → 1 ∈ ℝ
72 16 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → A ∈ ℝ
73 72 renegcld ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → − A ∈ ℝ
74 19 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → B ∈ ℝ
75 simprl ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → B ≠ 0
76 73 74 75 redivcld ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → − A B ∈ ℝ
77 1 2 prelrrx2 ⊢ 1 ∈ ℝ ∧ − A B ∈ ℝ → ⟨“ 1 − A B ”⟩ ∈ P
78 71 76 77 syl2anc ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → ⟨“ 1 − A B ”⟩ ∈ P
79 ax-1ne0 ⊢ 1 ≠ 0
80 79 neii ⊢ ¬ 1 = 0
81 10 80 2th ⊢ 0 = 0 ↔ ¬ 1 = 0
82 xor3 ⊢ ¬ 0 = 0 ↔ 1 = 0 ↔ 0 = 0 ↔ ¬ 1 = 0
83 81 82 mpbir ⊢ ¬ 0 = 0 ↔ 1 = 0
84 17 mulridd ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → A ⋅ 1 = A
85 84 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → A ⋅ 1 = A
86 17 negcld ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → − A ∈ ℂ
87 86 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → − A ∈ ℂ
88 20 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → B ∈ ℂ
89 87 88 75 divcan2d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → B ⁢ − A B = − A
90 85 89 oveq12d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → A ⋅ 1 + B ⁢ − A B = A + − A
91 17 negidd ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → A + − A = 0
92 91 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → A + − A = 0
93 90 92 eqtrd ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → A ⋅ 1 + B ⁢ − A B = 0
94 simprr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → C = 0
95 93 94 eqeq12d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → A ⋅ 1 + B ⁢ − A B = C ↔ 0 = 0
96 95 bibi1d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → A ⋅ 1 + B ⁢ − A B = C ↔ 1 = 0 ↔ 0 = 0 ↔ 1 = 0
97 83 96 mtbiri ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → ¬ A ⋅ 1 + B ⁢ − A B = C ↔ 1 = 0
98 70 78 97 rspcedvdw ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ B ≠ 0 ∧ C = 0 → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
99 98 expcom ⊢ B ≠ 0 ∧ C = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
100 99 ex ⊢ B ≠ 0 → C = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
101 53 100 sylbir ⊢ ¬ B = 0 → C = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
102 notnotb ⊢ B = 0 ↔ ¬ ¬ B = 0
103 nne ⊢ ¬ A ≠ 0 ↔ A = 0
104 103 bicomi ⊢ A = 0 ↔ ¬ A ≠ 0
105 1re ⊢ 1 ∈ ℝ
106 1 2 prelrrx2 ⊢ 1 ∈ ℝ ∧ 1 ∈ ℝ → ⟨“ 1 1 ”⟩ ∈ P
107 105 105 106 mp2an ⊢ ⟨“ 1 1 ”⟩ ∈ P
108 oveq1 ⊢ A = 0 → A ⋅ 1 = 0 ⋅ 1
109 108 adantl ⊢ B = 0 ∧ A = 0 → A ⋅ 1 = 0 ⋅ 1
110 ax-1cn ⊢ 1 ∈ ℂ
111 110 mul02i ⊢ 0 ⋅ 1 = 0
112 109 111 eqtrdi ⊢ B = 0 ∧ A = 0 → A ⋅ 1 = 0
113 oveq1 ⊢ B = 0 → B ⋅ 1 = 0 ⋅ 1
114 113 adantr ⊢ B = 0 ∧ A = 0 → B ⋅ 1 = 0 ⋅ 1
115 114 111 eqtrdi ⊢ B = 0 ∧ A = 0 → B ⋅ 1 = 0
116 112 115 oveq12d ⊢ B = 0 ∧ A = 0 → A ⋅ 1 + B ⋅ 1 = 0 + 0
117 116 23 eqtrdi ⊢ B = 0 ∧ A = 0 → A ⋅ 1 + B ⋅ 1 = 0
118 id ⊢ C = 0 → C = 0
119 117 118 eqeqan12d ⊢ B = 0 ∧ A = 0 ∧ C = 0 → A ⋅ 1 + B ⋅ 1 = C ↔ 0 = 0
120 119 bibi1d ⊢ B = 0 ∧ A = 0 ∧ C = 0 → A ⋅ 1 + B ⋅ 1 = C ↔ 1 = 0 ↔ 0 = 0 ↔ 1 = 0
121 83 120 mtbiri ⊢ B = 0 ∧ A = 0 ∧ C = 0 → ¬ A ⋅ 1 + B ⋅ 1 = C ↔ 1 = 0
122 fveq1 ⊢ p = ⟨“ 1 1 ”⟩ → p ⁡ 0 = ⟨“ 1 1 ”⟩ ⁡ 0
123 s2fv0 ⊢ 1 ∈ V → ⟨“ 1 1 ”⟩ ⁡ 0 = 1
124 55 123 ax-mp ⊢ ⟨“ 1 1 ”⟩ ⁡ 0 = 1
125 122 124 eqtrdi ⊢ p = ⟨“ 1 1 ”⟩ → p ⁡ 0 = 1
126 125 oveq2d ⊢ p = ⟨“ 1 1 ”⟩ → A ⁢ p ⁡ 0 = A ⋅ 1
127 fveq1 ⊢ p = ⟨“ 1 1 ”⟩ → p ⁡ 1 = ⟨“ 1 1 ”⟩ ⁡ 1
128 s2fv1 ⊢ 1 ∈ V → ⟨“ 1 1 ”⟩ ⁡ 1 = 1
129 55 128 ax-mp ⊢ ⟨“ 1 1 ”⟩ ⁡ 1 = 1
130 127 129 eqtrdi ⊢ p = ⟨“ 1 1 ”⟩ → p ⁡ 1 = 1
131 130 oveq2d ⊢ p = ⟨“ 1 1 ”⟩ → B ⁢ p ⁡ 1 = B ⋅ 1
132 126 131 oveq12d ⊢ p = ⟨“ 1 1 ”⟩ → A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = A ⋅ 1 + B ⋅ 1
133 132 eqeq1d ⊢ p = ⟨“ 1 1 ”⟩ → A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ A ⋅ 1 + B ⋅ 1 = C
134 125 eqeq1d ⊢ p = ⟨“ 1 1 ”⟩ → p ⁡ 0 = 0 ↔ 1 = 0
135 133 134 bibi12d ⊢ p = ⟨“ 1 1 ”⟩ → A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0 ↔ A ⋅ 1 + B ⋅ 1 = C ↔ 1 = 0
136 135 notbid ⊢ p = ⟨“ 1 1 ”⟩ → ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0 ↔ ¬ A ⋅ 1 + B ⋅ 1 = C ↔ 1 = 0
137 136 rspcev ⊢ ⟨“ 1 1 ”⟩ ∈ P ∧ ¬ A ⋅ 1 + B ⋅ 1 = C ↔ 1 = 0 → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
138 107 121 137 sylancr ⊢ B = 0 ∧ A = 0 ∧ C = 0 → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
139 138 a1d ⊢ B = 0 ∧ A = 0 ∧ C = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
140 139 ex ⊢ B = 0 ∧ A = 0 → C = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
141 102 104 140 syl2anbr ⊢ ¬ ¬ B = 0 ∧ ¬ A ≠ 0 → C = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
142 101 141 jaoi3 ⊢ ¬ B = 0 ∨ ¬ A ≠ 0 → C = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
143 142 orcoms ⊢ ¬ A ≠ 0 ∨ ¬ B = 0 → C = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
144 52 143 sylbi ⊢ ¬ A ≠ 0 ∧ B = 0 → C = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
145 144 com12 ⊢ C = 0 → ¬ A ≠ 0 ∧ B = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
146 51 145 sylbir ⊢ ¬ ¬ C = 0 → ¬ A ≠ 0 ∧ B = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
147 146 imp ⊢ ¬ ¬ C = 0 ∧ ¬ A ≠ 0 ∧ B = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
148 50 147 jaoi3 ⊢ ¬ C = 0 ∨ ¬ A ≠ 0 ∧ B = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
149 148 orcoms ⊢ ¬ A ≠ 0 ∧ B = 0 ∨ ¬ C = 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
150 149 com12 ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ¬ A ≠ 0 ∧ B = 0 ∨ ¬ C = 0 → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
151 3 150 biimtrid ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ¬ A ≠ 0 ∧ B = 0 ∧ C = 0 → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
152 rexnal ⊢ ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0 ↔ ¬ ∀ p ∈ P A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
153 151 152 imbitrdi ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ¬ A ≠ 0 ∧ B = 0 ∧ C = 0 → ¬ ∀ p ∈ P A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0
154 153 con4d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∀ p ∈ P A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0 → A ≠ 0 ∧ B = 0 ∧ C = 0
155 df-3an ⊢ A ≠ 0 ∧ B = 0 ∧ C = 0 ↔ A ≠ 0 ∧ B = 0 ∧ C = 0
156 154 155 imbitrrdi ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ → ∀ p ∈ P A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 0 = 0 → A ≠ 0 ∧ B = 0 ∧ C = 0