Metamath Proof Explorer


Theorem line2xlem

Description: Lemma for line2x . This proof is based on counterexamples for the following cases: 1. M =/= ( C / B ) : p = (0,C/B) (LHS of biconditional is true, RHS is false); 2. A =/= 0 /\ M = ( C / B ) : p = (1,C/B) (LHS of biconditional is false, RHS is true). (Contributed by AV, 4-Feb-2023)

Ref Expression
Hypotheses line2.i ⊢ I = 0 ..^ 2
line2.e ⊢ E = I
line2.p ⊢ P = ℝ I
line2.l ⊢ L = Line M ⁡ E
line2.g ⊢ G = p ∈ P | A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C
line2x.x ⊢ X = ⟨“ 0 M ”⟩
line2x.y ⊢ Y = ⟨“ 1 M ”⟩
Assertion line2xlem ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ∀ p ∈ P A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 1 = M → A = 0 ∧ M = C B

Proof

Step Hyp Ref Expression
1 line2.i ⊢ I = 0 ..^ 2
2 line2.e ⊢ E = I
3 line2.p ⊢ P = ℝ I
4 line2.l ⊢ L = Line M ⁡ E
5 line2.g ⊢ G = p ∈ P | A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C
6 line2x.x ⊢ X = ⟨“ 0 M ”⟩
7 line2x.y ⊢ Y = ⟨“ 1 M ”⟩
8 ianor ⊢ ¬ A = 0 ∧ M = C B ↔ ¬ A = 0 ∨ ¬ M = C B
9 df-ne ⊢ A ≠ 0 ↔ ¬ A = 0
10 df-ne ⊢ M ≠ C B ↔ ¬ M = C B
11 9 10 orbi12i ⊢ A ≠ 0 ∨ M ≠ C B ↔ ¬ A = 0 ∨ ¬ M = C B
12 8 11 bitr4i ⊢ ¬ A = 0 ∧ M = C B ↔ A ≠ 0 ∨ M ≠ C B
13 fveq1 ⊢ p = ⟨“ 0 C B ”⟩ → p ⁡ 0 = ⟨“ 0 C B ”⟩ ⁡ 0
14 13 oveq2d ⊢ p = ⟨“ 0 C B ”⟩ → A ⁢ p ⁡ 0 = A ⁢ ⟨“ 0 C B ”⟩ ⁡ 0
15 fveq1 ⊢ p = ⟨“ 0 C B ”⟩ → p ⁡ 1 = ⟨“ 0 C B ”⟩ ⁡ 1
16 15 oveq2d ⊢ p = ⟨“ 0 C B ”⟩ → B ⁢ p ⁡ 1 = B ⁢ ⟨“ 0 C B ”⟩ ⁡ 1
17 14 16 oveq12d ⊢ p = ⟨“ 0 C B ”⟩ → A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = A ⁢ ⟨“ 0 C B ”⟩ ⁡ 0 + B ⁢ ⟨“ 0 C B ”⟩ ⁡ 1
18 17 eqeq1d ⊢ p = ⟨“ 0 C B ”⟩ → A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ A ⁢ ⟨“ 0 C B ”⟩ ⁡ 0 + B ⁢ ⟨“ 0 C B ”⟩ ⁡ 1 = C
19 15 eqeq1d ⊢ p = ⟨“ 0 C B ”⟩ → p ⁡ 1 = M ↔ ⟨“ 0 C B ”⟩ ⁡ 1 = M
20 18 19 bibi12d ⊢ p = ⟨“ 0 C B ”⟩ → A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 1 = M ↔ A ⁢ ⟨“ 0 C B ”⟩ ⁡ 0 + B ⁢ ⟨“ 0 C B ”⟩ ⁡ 1 = C ↔ ⟨“ 0 C B ”⟩ ⁡ 1 = M
21 20 notbid ⊢ p = ⟨“ 0 C B ”⟩ → ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 1 = M ↔ ¬ A ⁢ ⟨“ 0 C B ”⟩ ⁡ 0 + B ⁢ ⟨“ 0 C B ”⟩ ⁡ 1 = C ↔ ⟨“ 0 C B ”⟩ ⁡ 1 = M
22 0red ⊢ M ≠ C B ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → 0 ∈ ℝ
23 simp3 ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ → C ∈ ℝ
24 23 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → C ∈ ℝ
25 simpl ⊢ B ∈ ℝ ∧ B ≠ 0 → B ∈ ℝ
26 25 3ad2ant2 ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ → B ∈ ℝ
27 26 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → B ∈ ℝ
28 simp2r ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ → B ≠ 0
29 28 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → B ≠ 0
30 24 27 29 redivcld ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → C B ∈ ℝ
31 30 adantl ⊢ M ≠ C B ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → C B ∈ ℝ
32 1 3 prelrrx2 ⊢ 0 ∈ ℝ ∧ C B ∈ ℝ → ⟨“ 0 C B ”⟩ ∈ P
33 22 31 32 syl2anc ⊢ M ≠ C B ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ⟨“ 0 C B ”⟩ ∈ P
34 id ⊢ M ≠ C B → M ≠ C B
35 34 necomd ⊢ M ≠ C B → C B ≠ M
36 35 neneqd ⊢ M ≠ C B → ¬ C B = M
37 36 a1d ⊢ M ≠ C B → C = C → ¬ C B = M
38 eqidd ⊢ ¬ C B = M → C = C
39 37 38 impbid1 ⊢ M ≠ C B → C = C ↔ ¬ C B = M
40 xor3 ⊢ ¬ C = C ↔ C B = M ↔ C = C ↔ ¬ C B = M
41 39 40 sylibr ⊢ M ≠ C B → ¬ C = C ↔ C B = M
42 41 adantr ⊢ M ≠ C B ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ¬ C = C ↔ C B = M
43 0red ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → 0 ∈ ℝ
44 s2fv0 ⊢ 0 ∈ ℝ → ⟨“ 0 C B ”⟩ ⁡ 0 = 0
45 43 44 syl ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ⟨“ 0 C B ”⟩ ⁡ 0 = 0
46 45 oveq2d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → A ⁢ ⟨“ 0 C B ”⟩ ⁡ 0 = A ⋅ 0
47 recn ⊢ A ∈ ℝ → A ∈ ℂ
48 47 mul01d ⊢ A ∈ ℝ → A ⋅ 0 = 0
49 48 3ad2ant1 ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ → A ⋅ 0 = 0
50 49 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → A ⋅ 0 = 0
51 46 50 eqtrd ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → A ⁢ ⟨“ 0 C B ”⟩ ⁡ 0 = 0
52 ovexd ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → C B ∈ V
53 s2fv1 ⊢ C B ∈ V → ⟨“ 0 C B ”⟩ ⁡ 1 = C B
54 52 53 syl ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ⟨“ 0 C B ”⟩ ⁡ 1 = C B
55 54 oveq2d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → B ⁢ ⟨“ 0 C B ”⟩ ⁡ 1 = B ⁢ C B
56 23 recnd ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ → C ∈ ℂ
57 56 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → C ∈ ℂ
58 25 recnd ⊢ B ∈ ℝ ∧ B ≠ 0 → B ∈ ℂ
59 58 3ad2ant2 ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ → B ∈ ℂ
60 59 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → B ∈ ℂ
61 57 60 29 divcan2d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → B ⁢ C B = C
62 55 61 eqtrd ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → B ⁢ ⟨“ 0 C B ”⟩ ⁡ 1 = C
63 51 62 oveq12d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → A ⁢ ⟨“ 0 C B ”⟩ ⁡ 0 + B ⁢ ⟨“ 0 C B ”⟩ ⁡ 1 = 0 + C
64 63 adantl ⊢ M ≠ C B ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → A ⁢ ⟨“ 0 C B ”⟩ ⁡ 0 + B ⁢ ⟨“ 0 C B ”⟩ ⁡ 1 = 0 + C
65 56 addlidd ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ → 0 + C = C
66 65 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → 0 + C = C
67 66 adantl ⊢ M ≠ C B ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → 0 + C = C
68 64 67 eqtrd ⊢ M ≠ C B ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → A ⁢ ⟨“ 0 C B ”⟩ ⁡ 0 + B ⁢ ⟨“ 0 C B ”⟩ ⁡ 1 = C
69 68 eqeq1d ⊢ M ≠ C B ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → A ⁢ ⟨“ 0 C B ”⟩ ⁡ 0 + B ⁢ ⟨“ 0 C B ”⟩ ⁡ 1 = C ↔ C = C
70 54 eqeq1d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ⟨“ 0 C B ”⟩ ⁡ 1 = M ↔ C B = M
71 70 adantl ⊢ M ≠ C B ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ⟨“ 0 C B ”⟩ ⁡ 1 = M ↔ C B = M
72 69 71 bibi12d ⊢ M ≠ C B ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → A ⁢ ⟨“ 0 C B ”⟩ ⁡ 0 + B ⁢ ⟨“ 0 C B ”⟩ ⁡ 1 = C ↔ ⟨“ 0 C B ”⟩ ⁡ 1 = M ↔ C = C ↔ C B = M
73 42 72 mtbird ⊢ M ≠ C B ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ¬ A ⁢ ⟨“ 0 C B ”⟩ ⁡ 0 + B ⁢ ⟨“ 0 C B ”⟩ ⁡ 1 = C ↔ ⟨“ 0 C B ”⟩ ⁡ 1 = M
74 21 33 73 rspcedvdw ⊢ M ≠ C B ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 1 = M
75 74 ex ⊢ M ≠ C B → A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 1 = M
76 nne ⊢ ¬ M ≠ C B ↔ M = C B
77 fveq1 ⊢ p = ⟨“ 1 C B ”⟩ → p ⁡ 0 = ⟨“ 1 C B ”⟩ ⁡ 0
78 77 oveq2d ⊢ p = ⟨“ 1 C B ”⟩ → A ⁢ p ⁡ 0 = A ⁢ ⟨“ 1 C B ”⟩ ⁡ 0
79 fveq1 ⊢ p = ⟨“ 1 C B ”⟩ → p ⁡ 1 = ⟨“ 1 C B ”⟩ ⁡ 1
80 79 oveq2d ⊢ p = ⟨“ 1 C B ”⟩ → B ⁢ p ⁡ 1 = B ⁢ ⟨“ 1 C B ”⟩ ⁡ 1
81 78 80 oveq12d ⊢ p = ⟨“ 1 C B ”⟩ → A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = A ⁢ ⟨“ 1 C B ”⟩ ⁡ 0 + B ⁢ ⟨“ 1 C B ”⟩ ⁡ 1
82 81 eqeq1d ⊢ p = ⟨“ 1 C B ”⟩ → A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ A ⁢ ⟨“ 1 C B ”⟩ ⁡ 0 + B ⁢ ⟨“ 1 C B ”⟩ ⁡ 1 = C
83 79 eqeq1d ⊢ p = ⟨“ 1 C B ”⟩ → p ⁡ 1 = M ↔ ⟨“ 1 C B ”⟩ ⁡ 1 = M
84 82 83 bibi12d ⊢ p = ⟨“ 1 C B ”⟩ → A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 1 = M ↔ A ⁢ ⟨“ 1 C B ”⟩ ⁡ 0 + B ⁢ ⟨“ 1 C B ”⟩ ⁡ 1 = C ↔ ⟨“ 1 C B ”⟩ ⁡ 1 = M
85 84 notbid ⊢ p = ⟨“ 1 C B ”⟩ → ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 1 = M ↔ ¬ A ⁢ ⟨“ 1 C B ”⟩ ⁡ 0 + B ⁢ ⟨“ 1 C B ”⟩ ⁡ 1 = C ↔ ⟨“ 1 C B ”⟩ ⁡ 1 = M
86 1red ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ → 1 ∈ ℝ
87 23 26 28 redivcld ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ → C B ∈ ℝ
88 86 87 jca ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ → 1 ∈ ℝ ∧ C B ∈ ℝ
89 88 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → 1 ∈ ℝ ∧ C B ∈ ℝ
90 1 3 prelrrx2 ⊢ 1 ∈ ℝ ∧ C B ∈ ℝ → ⟨“ 1 C B ”⟩ ∈ P
91 89 90 syl ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ⟨“ 1 C B ”⟩ ∈ P
92 91 adantl ⊢ M = C B ∧ A ≠ 0 ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ⟨“ 1 C B ”⟩ ∈ P
93 eqneqall ⊢ A = 0 → A ≠ 0 → ¬ C B = M
94 93 com12 ⊢ A ≠ 0 → A = 0 → ¬ C B = M
95 94 adantl ⊢ M = C B ∧ A ≠ 0 → A = 0 → ¬ C B = M
96 pm2.24 ⊢ C B = M → ¬ C B = M → A = 0
97 96 eqcoms ⊢ M = C B → ¬ C B = M → A = 0
98 97 adantr ⊢ M = C B ∧ A ≠ 0 → ¬ C B = M → A = 0
99 95 98 impbid ⊢ M = C B ∧ A ≠ 0 → A = 0 ↔ ¬ C B = M
100 xor3 ⊢ ¬ A = 0 ↔ C B = M ↔ A = 0 ↔ ¬ C B = M
101 99 100 sylibr ⊢ M = C B ∧ A ≠ 0 → ¬ A = 0 ↔ C B = M
102 101 adantr ⊢ M = C B ∧ A ≠ 0 ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ¬ A = 0 ↔ C B = M
103 simprl1 ⊢ M = C B ∧ A ≠ 0 ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → A ∈ ℝ
104 103 recnd ⊢ M = C B ∧ A ≠ 0 ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → A ∈ ℂ
105 24 adantl ⊢ M = C B ∧ A ≠ 0 ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → C ∈ ℝ
106 105 recnd ⊢ M = C B ∧ A ≠ 0 ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → C ∈ ℂ
107 104 106 addcomd ⊢ M = C B ∧ A ≠ 0 ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → A + C = C + A
108 107 eqeq1d ⊢ M = C B ∧ A ≠ 0 ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → A + C = C ↔ C + A = C
109 recn ⊢ C ∈ ℝ → C ∈ ℂ
110 47 109 anim12ci ⊢ A ∈ ℝ ∧ C ∈ ℝ → C ∈ ℂ ∧ A ∈ ℂ
111 110 3adant2 ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ → C ∈ ℂ ∧ A ∈ ℂ
112 111 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → C ∈ ℂ ∧ A ∈ ℂ
113 112 adantl ⊢ M = C B ∧ A ≠ 0 ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → C ∈ ℂ ∧ A ∈ ℂ
114 addid0 ⊢ C ∈ ℂ ∧ A ∈ ℂ → C + A = C ↔ A = 0
115 113 114 syl ⊢ M = C B ∧ A ≠ 0 ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → C + A = C ↔ A = 0
116 108 115 bitrd ⊢ M = C B ∧ A ≠ 0 ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → A + C = C ↔ A = 0
117 116 bibi1d ⊢ M = C B ∧ A ≠ 0 ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → A + C = C ↔ C B = M ↔ A = 0 ↔ C B = M
118 102 117 mtbird ⊢ M = C B ∧ A ≠ 0 ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ¬ A + C = C ↔ C B = M
119 1re ⊢ 1 ∈ ℝ
120 s2fv0 ⊢ 1 ∈ ℝ → ⟨“ 1 C B ”⟩ ⁡ 0 = 1
121 119 120 mp1i ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ⟨“ 1 C B ”⟩ ⁡ 0 = 1
122 121 oveq2d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → A ⁢ ⟨“ 1 C B ”⟩ ⁡ 0 = A ⋅ 1
123 ax-1rid ⊢ A ∈ ℝ → A ⋅ 1 = A
124 123 3ad2ant1 ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ → A ⋅ 1 = A
125 124 adantr ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → A ⋅ 1 = A
126 122 125 eqtrd ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → A ⁢ ⟨“ 1 C B ”⟩ ⁡ 0 = A
127 s2fv1 ⊢ C B ∈ V → ⟨“ 1 C B ”⟩ ⁡ 1 = C B
128 52 127 syl ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ⟨“ 1 C B ”⟩ ⁡ 1 = C B
129 128 oveq2d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → B ⁢ ⟨“ 1 C B ”⟩ ⁡ 1 = B ⁢ C B
130 129 61 eqtrd ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → B ⁢ ⟨“ 1 C B ”⟩ ⁡ 1 = C
131 126 130 oveq12d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → A ⁢ ⟨“ 1 C B ”⟩ ⁡ 0 + B ⁢ ⟨“ 1 C B ”⟩ ⁡ 1 = A + C
132 131 eqeq1d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → A ⁢ ⟨“ 1 C B ”⟩ ⁡ 0 + B ⁢ ⟨“ 1 C B ”⟩ ⁡ 1 = C ↔ A + C = C
133 128 eqeq1d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ⟨“ 1 C B ”⟩ ⁡ 1 = M ↔ C B = M
134 132 133 bibi12d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → A ⁢ ⟨“ 1 C B ”⟩ ⁡ 0 + B ⁢ ⟨“ 1 C B ”⟩ ⁡ 1 = C ↔ ⟨“ 1 C B ”⟩ ⁡ 1 = M ↔ A + C = C ↔ C B = M
135 134 notbid ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ¬ A ⁢ ⟨“ 1 C B ”⟩ ⁡ 0 + B ⁢ ⟨“ 1 C B ”⟩ ⁡ 1 = C ↔ ⟨“ 1 C B ”⟩ ⁡ 1 = M ↔ ¬ A + C = C ↔ C B = M
136 135 adantl ⊢ M = C B ∧ A ≠ 0 ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ¬ A ⁢ ⟨“ 1 C B ”⟩ ⁡ 0 + B ⁢ ⟨“ 1 C B ”⟩ ⁡ 1 = C ↔ ⟨“ 1 C B ”⟩ ⁡ 1 = M ↔ ¬ A + C = C ↔ C B = M
137 118 136 mpbird ⊢ M = C B ∧ A ≠ 0 ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ¬ A ⁢ ⟨“ 1 C B ”⟩ ⁡ 0 + B ⁢ ⟨“ 1 C B ”⟩ ⁡ 1 = C ↔ ⟨“ 1 C B ”⟩ ⁡ 1 = M
138 85 92 137 rspcedvdw ⊢ M = C B ∧ A ≠ 0 ∧ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 1 = M
139 138 ex ⊢ M = C B ∧ A ≠ 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 1 = M
140 76 139 sylanb ⊢ ¬ M ≠ C B ∧ A ≠ 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 1 = M
141 75 140 jaoi3 ⊢ M ≠ C B ∨ A ≠ 0 → A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 1 = M
142 141 orcoms ⊢ A ≠ 0 ∨ M ≠ C B → A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 1 = M
143 142 com12 ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → A ≠ 0 ∨ M ≠ C B → ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 1 = M
144 rexnal ⊢ ∃ p ∈ P ¬ A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 1 = M ↔ ¬ ∀ p ∈ P A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 1 = M
145 143 144 imbitrdi ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → A ≠ 0 ∨ M ≠ C B → ¬ ∀ p ∈ P A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 1 = M
146 12 145 biimtrid ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ¬ A = 0 ∧ M = C B → ¬ ∀ p ∈ P A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 1 = M
147 146 con4d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ B ≠ 0 ∧ C ∈ ℝ ∧ M ∈ ℝ → ∀ p ∈ P A ⁢ p ⁡ 0 + B ⁢ p ⁡ 1 = C ↔ p ⁡ 1 = M → A = 0 ∧ M = C B