Theorem add4

Description: Rearrangement of 4 terms in a sum. (Contributed by NM, 13-Nov-1999) (Proof shortened by Andrew Salmon, 22-Oct-2011)

	Ref	Expression
Assertion	add4	⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ) ) → ( ( 𝐴 + 𝐵 ) + ( 𝐶 + 𝐷 ) ) = ( ( 𝐴 + 𝐶 ) + ( 𝐵 + 𝐷 ) ) )

Proof

Step	Hyp	Ref	Expression
1		add12	⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ) → ( 𝐵 + ( 𝐶 + 𝐷 ) ) = ( 𝐶 + ( 𝐵 + 𝐷 ) ) )
2	1	3expb	⊢ ( ( 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ) ) → ( 𝐵 + ( 𝐶 + 𝐷 ) ) = ( 𝐶 + ( 𝐵 + 𝐷 ) ) )
3	2	oveq2d	⊢ ( ( 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ) ) → ( 𝐴 + ( 𝐵 + ( 𝐶 + 𝐷 ) ) ) = ( 𝐴 + ( 𝐶 + ( 𝐵 + 𝐷 ) ) ) )
4	3	adantll	⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ) ) → ( 𝐴 + ( 𝐵 + ( 𝐶 + 𝐷 ) ) ) = ( 𝐴 + ( 𝐶 + ( 𝐵 + 𝐷 ) ) ) )
5		addcl	⊢ ( ( 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ) → ( 𝐶 + 𝐷 ) ∈ ℂ )
6		addass	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 + 𝐷 ) ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) + ( 𝐶 + 𝐷 ) ) = ( 𝐴 + ( 𝐵 + ( 𝐶 + 𝐷 ) ) ) )
7	6	3expa	⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝐶 + 𝐷 ) ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) + ( 𝐶 + 𝐷 ) ) = ( 𝐴 + ( 𝐵 + ( 𝐶 + 𝐷 ) ) ) )
8	5 7	sylan2	⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ) ) → ( ( 𝐴 + 𝐵 ) + ( 𝐶 + 𝐷 ) ) = ( 𝐴 + ( 𝐵 + ( 𝐶 + 𝐷 ) ) ) )
9		addcl	⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐷 ∈ ℂ ) → ( 𝐵 + 𝐷 ) ∈ ℂ )
10		addass	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ ( 𝐵 + 𝐷 ) ∈ ℂ ) → ( ( 𝐴 + 𝐶 ) + ( 𝐵 + 𝐷 ) ) = ( 𝐴 + ( 𝐶 + ( 𝐵 + 𝐷 ) ) ) )
11	10	3expa	⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ∧ ( 𝐵 + 𝐷 ) ∈ ℂ ) → ( ( 𝐴 + 𝐶 ) + ( 𝐵 + 𝐷 ) ) = ( 𝐴 + ( 𝐶 + ( 𝐵 + 𝐷 ) ) ) )
12	9 11	sylan2	⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ∧ ( 𝐵 ∈ ℂ ∧ 𝐷 ∈ ℂ ) ) → ( ( 𝐴 + 𝐶 ) + ( 𝐵 + 𝐷 ) ) = ( 𝐴 + ( 𝐶 + ( 𝐵 + 𝐷 ) ) ) )
13	12	an4s	⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ) ) → ( ( 𝐴 + 𝐶 ) + ( 𝐵 + 𝐷 ) ) = ( 𝐴 + ( 𝐶 + ( 𝐵 + 𝐷 ) ) ) )
14	4 8 13	3eqtr4d	⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ) ) → ( ( 𝐴 + 𝐵 ) + ( 𝐶 + 𝐷 ) ) = ( ( 𝐴 + 𝐶 ) + ( 𝐵 + 𝐷 ) ) )