Metamath Proof Explorer


Theorem remulcand

Description: Commuted version of remulcan2d without ax-mulcom . (Contributed by SN, 21-Feb-2024)

Ref Expression
Hypotheses remulcand.1 ⊢ φ → A ∈ ℝ
remulcand.2 ⊢ φ → B ∈ ℝ
remulcand.3 ⊢ φ → C ∈ ℝ
remulcand.4 ⊢ φ → C ≠ 0
Assertion remulcand ⊢ φ → C ⁢ A = C ⁢ B ↔ A = B

Proof

Step Hyp Ref Expression
1 remulcand.1 ⊢ φ → A ∈ ℝ
2 remulcand.2 ⊢ φ → B ∈ ℝ
3 remulcand.3 ⊢ φ → C ∈ ℝ
4 remulcand.4 ⊢ φ → C ≠ 0
5 ax-rrecex ⊢ C ∈ ℝ ∧ C ≠ 0 → ∃ x ∈ ℝ C ⁢ x = 1
6 3 4 5 syl2anc ⊢ φ → ∃ x ∈ ℝ C ⁢ x = 1
7 3 adantr ⊢ φ ∧ x ∈ ℝ → C ∈ ℝ
8 7 adantr ⊢ φ ∧ x ∈ ℝ ∧ C ⁢ x = 1 → C ∈ ℝ
9 simplr ⊢ φ ∧ x ∈ ℝ ∧ C ⁢ x = 1 → x ∈ ℝ
10 simpr ⊢ φ ∧ x ∈ ℝ ∧ C ⁢ x = 1 → C ⁢ x = 1
11 8 9 10 remulinvcom ⊢ φ ∧ x ∈ ℝ ∧ C ⁢ x = 1 → x ⁢ C = 1
12 11 ex ⊢ φ ∧ x ∈ ℝ → C ⁢ x = 1 → x ⁢ C = 1
13 oveq2 ⊢ C ⁢ A = C ⁢ B → x ⁢ C ⁢ A = x ⁢ C ⁢ B
14 13 3ad2ant3 ⊢ φ ∧ x ∈ ℝ ∧ x ⁢ C = 1 ∧ C ⁢ A = C ⁢ B → x ⁢ C ⁢ A = x ⁢ C ⁢ B
15 simp2 ⊢ φ ∧ x ∈ ℝ ∧ x ⁢ C = 1 ∧ C ⁢ A = C ⁢ B → x ⁢ C = 1
16 15 oveq1d ⊢ φ ∧ x ∈ ℝ ∧ x ⁢ C = 1 ∧ C ⁢ A = C ⁢ B → x ⁢ C ⁢ A = 1 ⁢ A
17 simp1r ⊢ φ ∧ x ∈ ℝ ∧ x ⁢ C = 1 ∧ C ⁢ A = C ⁢ B → x ∈ ℝ
18 17 recnd ⊢ φ ∧ x ∈ ℝ ∧ x ⁢ C = 1 ∧ C ⁢ A = C ⁢ B → x ∈ ℂ
19 7 3ad2ant1 ⊢ φ ∧ x ∈ ℝ ∧ x ⁢ C = 1 ∧ C ⁢ A = C ⁢ B → C ∈ ℝ
20 19 recnd ⊢ φ ∧ x ∈ ℝ ∧ x ⁢ C = 1 ∧ C ⁢ A = C ⁢ B → C ∈ ℂ
21 simp1l ⊢ φ ∧ x ∈ ℝ ∧ x ⁢ C = 1 ∧ C ⁢ A = C ⁢ B → φ
22 21 1 syl ⊢ φ ∧ x ∈ ℝ ∧ x ⁢ C = 1 ∧ C ⁢ A = C ⁢ B → A ∈ ℝ
23 22 recnd ⊢ φ ∧ x ∈ ℝ ∧ x ⁢ C = 1 ∧ C ⁢ A = C ⁢ B → A ∈ ℂ
24 18 20 23 mulassd ⊢ φ ∧ x ∈ ℝ ∧ x ⁢ C = 1 ∧ C ⁢ A = C ⁢ B → x ⁢ C ⁢ A = x ⁢ C ⁢ A
25 remullid ⊢ A ∈ ℝ → 1 ⁢ A = A
26 22 25 syl ⊢ φ ∧ x ∈ ℝ ∧ x ⁢ C = 1 ∧ C ⁢ A = C ⁢ B → 1 ⁢ A = A
27 16 24 26 3eqtr3d ⊢ φ ∧ x ∈ ℝ ∧ x ⁢ C = 1 ∧ C ⁢ A = C ⁢ B → x ⁢ C ⁢ A = A
28 15 oveq1d ⊢ φ ∧ x ∈ ℝ ∧ x ⁢ C = 1 ∧ C ⁢ A = C ⁢ B → x ⁢ C ⁢ B = 1 ⁢ B
29 21 2 syl ⊢ φ ∧ x ∈ ℝ ∧ x ⁢ C = 1 ∧ C ⁢ A = C ⁢ B → B ∈ ℝ
30 29 recnd ⊢ φ ∧ x ∈ ℝ ∧ x ⁢ C = 1 ∧ C ⁢ A = C ⁢ B → B ∈ ℂ
31 18 20 30 mulassd ⊢ φ ∧ x ∈ ℝ ∧ x ⁢ C = 1 ∧ C ⁢ A = C ⁢ B → x ⁢ C ⁢ B = x ⁢ C ⁢ B
32 remullid ⊢ B ∈ ℝ → 1 ⁢ B = B
33 29 32 syl ⊢ φ ∧ x ∈ ℝ ∧ x ⁢ C = 1 ∧ C ⁢ A = C ⁢ B → 1 ⁢ B = B
34 28 31 33 3eqtr3d ⊢ φ ∧ x ∈ ℝ ∧ x ⁢ C = 1 ∧ C ⁢ A = C ⁢ B → x ⁢ C ⁢ B = B
35 14 27 34 3eqtr3d ⊢ φ ∧ x ∈ ℝ ∧ x ⁢ C = 1 ∧ C ⁢ A = C ⁢ B → A = B
36 35 3exp ⊢ φ ∧ x ∈ ℝ → x ⁢ C = 1 → C ⁢ A = C ⁢ B → A = B
37 12 36 syld ⊢ φ ∧ x ∈ ℝ → C ⁢ x = 1 → C ⁢ A = C ⁢ B → A = B
38 37 impr ⊢ φ ∧ x ∈ ℝ ∧ C ⁢ x = 1 → C ⁢ A = C ⁢ B → A = B
39 6 38 rexlimddv ⊢ φ → C ⁢ A = C ⁢ B → A = B
40 oveq2 ⊢ A = B → C ⁢ A = C ⁢ B
41 39 40 impbid1 ⊢ φ → C ⁢ A = C ⁢ B ↔ A = B