Metamath Proof Explorer


Theorem ichexmpl2

Description: Example for interchangeable setvar variables in an arithmetic expression. (Contributed by AV, 31-Jul-2023)

Ref Expression
Assertion ichexmpl2 ⊢ a ⇄ b a ∈ ℂ ∧ b ∈ ℂ ∧ c ∈ ℂ → a 2 + b 2 = c 2

Proof

Step Hyp Ref Expression
1 eleq1w ⊢ a = t → a ∈ ℂ ↔ t ∈ ℂ
2 1 3anbi1d ⊢ a = t → a ∈ ℂ ∧ b ∈ ℂ ∧ c ∈ ℂ ↔ t ∈ ℂ ∧ b ∈ ℂ ∧ c ∈ ℂ
3 oveq1 ⊢ a = t → a 2 = t 2
4 3 oveq1d ⊢ a = t → a 2 + b 2 = t 2 + b 2
5 4 eqeq1d ⊢ a = t → a 2 + b 2 = c 2 ↔ t 2 + b 2 = c 2
6 2 5 imbi12d ⊢ a = t → a ∈ ℂ ∧ b ∈ ℂ ∧ c ∈ ℂ → a 2 + b 2 = c 2 ↔ t ∈ ℂ ∧ b ∈ ℂ ∧ c ∈ ℂ → t 2 + b 2 = c 2
7 eleq1w ⊢ b = a → b ∈ ℂ ↔ a ∈ ℂ
8 7 3anbi2d ⊢ b = a → t ∈ ℂ ∧ b ∈ ℂ ∧ c ∈ ℂ ↔ t ∈ ℂ ∧ a ∈ ℂ ∧ c ∈ ℂ
9 oveq1 ⊢ b = a → b 2 = a 2
10 9 oveq2d ⊢ b = a → t 2 + b 2 = t 2 + a 2
11 10 eqeq1d ⊢ b = a → t 2 + b 2 = c 2 ↔ t 2 + a 2 = c 2
12 8 11 imbi12d ⊢ b = a → t ∈ ℂ ∧ b ∈ ℂ ∧ c ∈ ℂ → t 2 + b 2 = c 2 ↔ t ∈ ℂ ∧ a ∈ ℂ ∧ c ∈ ℂ → t 2 + a 2 = c 2
13 eleq1w ⊢ t = b → t ∈ ℂ ↔ b ∈ ℂ
14 13 3anbi1d ⊢ t = b → t ∈ ℂ ∧ a ∈ ℂ ∧ c ∈ ℂ ↔ b ∈ ℂ ∧ a ∈ ℂ ∧ c ∈ ℂ
15 oveq1 ⊢ t = b → t 2 = b 2
16 15 oveq1d ⊢ t = b → t 2 + a 2 = b 2 + a 2
17 16 eqeq1d ⊢ t = b → t 2 + a 2 = c 2 ↔ b 2 + a 2 = c 2
18 14 17 imbi12d ⊢ t = b → t ∈ ℂ ∧ a ∈ ℂ ∧ c ∈ ℂ → t 2 + a 2 = c 2 ↔ b ∈ ℂ ∧ a ∈ ℂ ∧ c ∈ ℂ → b 2 + a 2 = c 2
19 3ancoma ⊢ b ∈ ℂ ∧ a ∈ ℂ ∧ c ∈ ℂ ↔ a ∈ ℂ ∧ b ∈ ℂ ∧ c ∈ ℂ
20 19 imbi1i ⊢ b ∈ ℂ ∧ a ∈ ℂ ∧ c ∈ ℂ → b 2 + a 2 = c 2 ↔ a ∈ ℂ ∧ b ∈ ℂ ∧ c ∈ ℂ → b 2 + a 2 = c 2
21 sqcl ⊢ b ∈ ℂ → b 2 ∈ ℂ
22 21 3ad2ant2 ⊢ a ∈ ℂ ∧ b ∈ ℂ ∧ c ∈ ℂ → b 2 ∈ ℂ
23 sqcl ⊢ a ∈ ℂ → a 2 ∈ ℂ
24 23 3ad2ant1 ⊢ a ∈ ℂ ∧ b ∈ ℂ ∧ c ∈ ℂ → a 2 ∈ ℂ
25 22 24 addcomd ⊢ a ∈ ℂ ∧ b ∈ ℂ ∧ c ∈ ℂ → b 2 + a 2 = a 2 + b 2
26 25 eqeq1d ⊢ a ∈ ℂ ∧ b ∈ ℂ ∧ c ∈ ℂ → b 2 + a 2 = c 2 ↔ a 2 + b 2 = c 2
27 26 pm5.74i ⊢ a ∈ ℂ ∧ b ∈ ℂ ∧ c ∈ ℂ → b 2 + a 2 = c 2 ↔ a ∈ ℂ ∧ b ∈ ℂ ∧ c ∈ ℂ → a 2 + b 2 = c 2
28 20 27 bitri ⊢ b ∈ ℂ ∧ a ∈ ℂ ∧ c ∈ ℂ → b 2 + a 2 = c 2 ↔ a ∈ ℂ ∧ b ∈ ℂ ∧ c ∈ ℂ → a 2 + b 2 = c 2
29 18 28 bitrdi ⊢ t = b → t ∈ ℂ ∧ a ∈ ℂ ∧ c ∈ ℂ → t 2 + a 2 = c 2 ↔ a ∈ ℂ ∧ b ∈ ℂ ∧ c ∈ ℂ → a 2 + b 2 = c 2
30 6 12 29 ichcircshi ⊢ a ⇄ b a ∈ ℂ ∧ b ∈ ℂ ∧ c ∈ ℂ → a 2 + b 2 = c 2