Description: Transfer two terms of a subtraction in an equality. (Contributed by Thierry Arnoux, 2-Feb-2020)
Ref | Expression | ||
---|---|---|---|
Hypotheses | subeqxfrd.a | ⊢ ( 𝜑 → 𝐴 ∈ ℂ ) | |
subeqxfrd.b | ⊢ ( 𝜑 → 𝐵 ∈ ℂ ) | ||
subeqxfrd.c | ⊢ ( 𝜑 → 𝐶 ∈ ℂ ) | ||
subeqxfrd.d | ⊢ ( 𝜑 → 𝐷 ∈ ℂ ) | ||
subeqxfrd.1 | ⊢ ( 𝜑 → ( 𝐴 − 𝐵 ) = ( 𝐶 − 𝐷 ) ) | ||
Assertion | subeqxfrd | ⊢ ( 𝜑 → ( 𝐴 − 𝐶 ) = ( 𝐵 − 𝐷 ) ) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | subeqxfrd.a | ⊢ ( 𝜑 → 𝐴 ∈ ℂ ) | |
2 | subeqxfrd.b | ⊢ ( 𝜑 → 𝐵 ∈ ℂ ) | |
3 | subeqxfrd.c | ⊢ ( 𝜑 → 𝐶 ∈ ℂ ) | |
4 | subeqxfrd.d | ⊢ ( 𝜑 → 𝐷 ∈ ℂ ) | |
5 | subeqxfrd.1 | ⊢ ( 𝜑 → ( 𝐴 − 𝐵 ) = ( 𝐶 − 𝐷 ) ) | |
6 | 5 | oveq1d | ⊢ ( 𝜑 → ( ( 𝐴 − 𝐵 ) + ( 𝐵 − 𝐶 ) ) = ( ( 𝐶 − 𝐷 ) + ( 𝐵 − 𝐶 ) ) ) |
7 | 1 2 3 | npncand | ⊢ ( 𝜑 → ( ( 𝐴 − 𝐵 ) + ( 𝐵 − 𝐶 ) ) = ( 𝐴 − 𝐶 ) ) |
8 | 3 4 2 | npncan3d | ⊢ ( 𝜑 → ( ( 𝐶 − 𝐷 ) + ( 𝐵 − 𝐶 ) ) = ( 𝐵 − 𝐷 ) ) |
9 | 6 7 8 | 3eqtr3d | ⊢ ( 𝜑 → ( 𝐴 − 𝐶 ) = ( 𝐵 − 𝐷 ) ) |