Description: Swapping order of subtraction doesn't change the absolute value. (Contributed by NM, 1-Oct-1999) (Proof shortened by Mario Carneiro, 29-May-2016)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | abssub | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( abs ‘ ( 𝐴 − 𝐵 ) ) = ( abs ‘ ( 𝐵 − 𝐴 ) ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | subcl | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 − 𝐵 ) ∈ ℂ ) | |
| 2 | absneg | ⊢ ( ( 𝐴 − 𝐵 ) ∈ ℂ → ( abs ‘ - ( 𝐴 − 𝐵 ) ) = ( abs ‘ ( 𝐴 − 𝐵 ) ) ) | |
| 3 | 1 2 | syl | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( abs ‘ - ( 𝐴 − 𝐵 ) ) = ( abs ‘ ( 𝐴 − 𝐵 ) ) ) |
| 4 | negsubdi2 | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → - ( 𝐴 − 𝐵 ) = ( 𝐵 − 𝐴 ) ) | |
| 5 | 4 | fveq2d | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( abs ‘ - ( 𝐴 − 𝐵 ) ) = ( abs ‘ ( 𝐵 − 𝐴 ) ) ) |
| 6 | 3 5 | eqtr3d | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( abs ‘ ( 𝐴 − 𝐵 ) ) = ( abs ‘ ( 𝐵 − 𝐴 ) ) ) |