Metamath Proof Explorer


Theorem nadd4

Description: Rearragement of terms in a quadruple sum. (Contributed by Scott Fenton, 5-Feb-2025)

Ref Expression
Assertion nadd4 ( ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) ∧ ( 𝐶 ∈ On ∧ 𝐷 ∈ On ) ) → ( ( 𝐴 +no 𝐵 ) +no ( 𝐶 +no 𝐷 ) ) = ( ( 𝐴 +no 𝐶 ) +no ( 𝐵 +no 𝐷 ) ) )

Proof

Step Hyp Ref Expression
1 nadd32 ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ∧ 𝐶 ∈ On ) → ( ( 𝐴 +no 𝐵 ) +no 𝐶 ) = ( ( 𝐴 +no 𝐶 ) +no 𝐵 ) )
2 1 3expa ( ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) ∧ 𝐶 ∈ On ) → ( ( 𝐴 +no 𝐵 ) +no 𝐶 ) = ( ( 𝐴 +no 𝐶 ) +no 𝐵 ) )
3 2 adantrr ( ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) ∧ ( 𝐶 ∈ On ∧ 𝐷 ∈ On ) ) → ( ( 𝐴 +no 𝐵 ) +no 𝐶 ) = ( ( 𝐴 +no 𝐶 ) +no 𝐵 ) )
4 3 oveq1d ( ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) ∧ ( 𝐶 ∈ On ∧ 𝐷 ∈ On ) ) → ( ( ( 𝐴 +no 𝐵 ) +no 𝐶 ) +no 𝐷 ) = ( ( ( 𝐴 +no 𝐶 ) +no 𝐵 ) +no 𝐷 ) )
5 naddcl ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) → ( 𝐴 +no 𝐵 ) ∈ On )
6 5 adantr ( ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) ∧ ( 𝐶 ∈ On ∧ 𝐷 ∈ On ) ) → ( 𝐴 +no 𝐵 ) ∈ On )
7 simprl ( ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) ∧ ( 𝐶 ∈ On ∧ 𝐷 ∈ On ) ) → 𝐶 ∈ On )
8 simprr ( ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) ∧ ( 𝐶 ∈ On ∧ 𝐷 ∈ On ) ) → 𝐷 ∈ On )
9 naddass ( ( ( 𝐴 +no 𝐵 ) ∈ On ∧ 𝐶 ∈ On ∧ 𝐷 ∈ On ) → ( ( ( 𝐴 +no 𝐵 ) +no 𝐶 ) +no 𝐷 ) = ( ( 𝐴 +no 𝐵 ) +no ( 𝐶 +no 𝐷 ) ) )
10 6 7 8 9 syl3anc ( ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) ∧ ( 𝐶 ∈ On ∧ 𝐷 ∈ On ) ) → ( ( ( 𝐴 +no 𝐵 ) +no 𝐶 ) +no 𝐷 ) = ( ( 𝐴 +no 𝐵 ) +no ( 𝐶 +no 𝐷 ) ) )
11 naddcl ( ( 𝐴 ∈ On ∧ 𝐶 ∈ On ) → ( 𝐴 +no 𝐶 ) ∈ On )
12 11 ad2ant2r ( ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) ∧ ( 𝐶 ∈ On ∧ 𝐷 ∈ On ) ) → ( 𝐴 +no 𝐶 ) ∈ On )
13 simplr ( ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) ∧ ( 𝐶 ∈ On ∧ 𝐷 ∈ On ) ) → 𝐵 ∈ On )
14 naddass ( ( ( 𝐴 +no 𝐶 ) ∈ On ∧ 𝐵 ∈ On ∧ 𝐷 ∈ On ) → ( ( ( 𝐴 +no 𝐶 ) +no 𝐵 ) +no 𝐷 ) = ( ( 𝐴 +no 𝐶 ) +no ( 𝐵 +no 𝐷 ) ) )
15 12 13 8 14 syl3anc ( ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) ∧ ( 𝐶 ∈ On ∧ 𝐷 ∈ On ) ) → ( ( ( 𝐴 +no 𝐶 ) +no 𝐵 ) +no 𝐷 ) = ( ( 𝐴 +no 𝐶 ) +no ( 𝐵 +no 𝐷 ) ) )
16 4 10 15 3eqtr3d ( ( ( 𝐴 ∈ On ∧ 𝐵 ∈ On ) ∧ ( 𝐶 ∈ On ∧ 𝐷 ∈ On ) ) → ( ( 𝐴 +no 𝐵 ) +no ( 𝐶 +no 𝐷 ) ) = ( ( 𝐴 +no 𝐶 ) +no ( 𝐵 +no 𝐷 ) ) )