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Metamath Proof Explorer

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 ) ) )