Theorem subaddeqd

Description: Transfer two terms of a subtraction to an addition in an equality. (Contributed by Thierry Arnoux, 2-Feb-2020)

	Ref	Expression
Hypotheses	subaddeqd.a	⊢ ( 𝜑 → 𝐴 ∈ ℂ )
	subaddeqd.b	⊢ ( 𝜑 → 𝐵 ∈ ℂ )
	subaddeqd.c	⊢ ( 𝜑 → 𝐶 ∈ ℂ )
	subaddeqd.d	⊢ ( 𝜑 → 𝐷 ∈ ℂ )
	subaddeqd.1	⊢ ( 𝜑 → ( 𝐴 + 𝐵 ) = ( 𝐶 + 𝐷 ) )
Assertion	subaddeqd	⊢ ( 𝜑 → ( 𝐴 − 𝐷 ) = ( 𝐶 − 𝐵 ) )

Proof

Step	Hyp	Ref	Expression
1		subaddeqd.a	⊢ ( 𝜑 → 𝐴 ∈ ℂ )
2		subaddeqd.b	⊢ ( 𝜑 → 𝐵 ∈ ℂ )
3		subaddeqd.c	⊢ ( 𝜑 → 𝐶 ∈ ℂ )
4		subaddeqd.d	⊢ ( 𝜑 → 𝐷 ∈ ℂ )
5		subaddeqd.1	⊢ ( 𝜑 → ( 𝐴 + 𝐵 ) = ( 𝐶 + 𝐷 ) )
6	5	oveq1d	⊢ ( 𝜑 → ( ( 𝐴 + 𝐵 ) − ( 𝐷 + 𝐵 ) ) = ( ( 𝐶 + 𝐷 ) − ( 𝐷 + 𝐵 ) ) )
7	3 4	addcomd	⊢ ( 𝜑 → ( 𝐶 + 𝐷 ) = ( 𝐷 + 𝐶 ) )
8	7	oveq1d	⊢ ( 𝜑 → ( ( 𝐶 + 𝐷 ) − ( 𝐷 + 𝐵 ) ) = ( ( 𝐷 + 𝐶 ) − ( 𝐷 + 𝐵 ) ) )
9	6 8	eqtrd	⊢ ( 𝜑 → ( ( 𝐴 + 𝐵 ) − ( 𝐷 + 𝐵 ) ) = ( ( 𝐷 + 𝐶 ) − ( 𝐷 + 𝐵 ) ) )
10	1 4 2	pnpcan2d	⊢ ( 𝜑 → ( ( 𝐴 + 𝐵 ) − ( 𝐷 + 𝐵 ) ) = ( 𝐴 − 𝐷 ) )
11	4 3 2	pnpcand	⊢ ( 𝜑 → ( ( 𝐷 + 𝐶 ) − ( 𝐷 + 𝐵 ) ) = ( 𝐶 − 𝐵 ) )
12	9 10 11	3eqtr3d	⊢ ( 𝜑 → ( 𝐴 − 𝐷 ) = ( 𝐶 − 𝐵 ) )