Metamath Proof Explorer
Description: Rearrangement of 4 terms in a subtraction. (Contributed by Mario
Carneiro, 27-May-2016)
|
|
Ref |
Expression |
|
Hypotheses |
negidd.1 |
⊢ ( 𝜑 → 𝐴 ∈ ℂ ) |
|
|
pncand.2 |
⊢ ( 𝜑 → 𝐵 ∈ ℂ ) |
|
|
subaddd.3 |
⊢ ( 𝜑 → 𝐶 ∈ ℂ ) |
|
|
addsub4d.4 |
⊢ ( 𝜑 → 𝐷 ∈ ℂ ) |
|
Assertion |
sub4d |
⊢ ( 𝜑 → ( ( 𝐴 − 𝐵 ) − ( 𝐶 − 𝐷 ) ) = ( ( 𝐴 − 𝐶 ) − ( 𝐵 − 𝐷 ) ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
negidd.1 |
⊢ ( 𝜑 → 𝐴 ∈ ℂ ) |
2 |
|
pncand.2 |
⊢ ( 𝜑 → 𝐵 ∈ ℂ ) |
3 |
|
subaddd.3 |
⊢ ( 𝜑 → 𝐶 ∈ ℂ ) |
4 |
|
addsub4d.4 |
⊢ ( 𝜑 → 𝐷 ∈ ℂ ) |
5 |
|
sub4 |
⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ) ) → ( ( 𝐴 − 𝐵 ) − ( 𝐶 − 𝐷 ) ) = ( ( 𝐴 − 𝐶 ) − ( 𝐵 − 𝐷 ) ) ) |
6 |
1 2 3 4 5
|
syl22anc |
⊢ ( 𝜑 → ( ( 𝐴 − 𝐵 ) − ( 𝐶 − 𝐷 ) ) = ( ( 𝐴 − 𝐶 ) − ( 𝐵 − 𝐷 ) ) ) |