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Metamath Proof Explorer

Theorem subadd4b

Description: Rearrangement of 4 terms in a mixed addition and subtraction. (Contributed by Glauco Siliprandi, 11-Dec-2019)

Ref Expression
Hypotheses subadd4b.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
subadd4b.2 ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
subadd4b.3 ⊢ ( 𝜑 → 𝐶 ∈ ℂ )
subadd4b.4 ⊢ ( 𝜑 → 𝐷 ∈ ℂ )
Assertion subadd4b ( 𝜑 → ( ( 𝐴 − 𝐵 ) + ( 𝐶 − 𝐷 ) ) = ( ( 𝐴 − 𝐷 ) + ( 𝐶 − 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 subadd4b.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
2 subadd4b.2 ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
3 subadd4b.3 ⊢ ( 𝜑 → 𝐶 ∈ ℂ )
4 subadd4b.4 ⊢ ( 𝜑 → 𝐷 ∈ ℂ )
5 1 2 4 3 subadd4d ⊢ ( 𝜑 → ( ( 𝐴 − 𝐵 ) − ( 𝐷 − 𝐶 ) ) = ( ( 𝐴 + 𝐶 ) − ( 𝐵 + 𝐷 ) ) )
6 1 2 subcld ⊢ ( 𝜑 → ( 𝐴 − 𝐵 ) ∈ ℂ )
7 6 4 3 subsub2d ⊢ ( 𝜑 → ( ( 𝐴 − 𝐵 ) − ( 𝐷 − 𝐶 ) ) = ( ( 𝐴 − 𝐵 ) + ( 𝐶 − 𝐷 ) ) )
8 2 4 addcomd ⊢ ( 𝜑 → ( 𝐵 + 𝐷 ) = ( 𝐷 + 𝐵 ) )
9 8 oveq2d ⊢ ( 𝜑 → ( ( 𝐴 + 𝐶 ) − ( 𝐵 + 𝐷 ) ) = ( ( 𝐴 + 𝐶 ) − ( 𝐷 + 𝐵 ) ) )
10 1 3 4 2 addsub4d ⊢ ( 𝜑 → ( ( 𝐴 + 𝐶 ) − ( 𝐷 + 𝐵 ) ) = ( ( 𝐴 − 𝐷 ) + ( 𝐶 − 𝐵 ) ) )
11 9 10 eqtrd ⊢ ( 𝜑 → ( ( 𝐴 + 𝐶 ) − ( 𝐵 + 𝐷 ) ) = ( ( 𝐴 − 𝐷 ) + ( 𝐶 − 𝐵 ) ) )
12 5 7 11 3eqtr3d ⊢ ( 𝜑 → ( ( 𝐴 − 𝐵 ) + ( 𝐶 − 𝐷 ) ) = ( ( 𝐴 − 𝐷 ) + ( 𝐶 − 𝐵 ) ) )