Metamath Proof Explorer


Theorem subsubs4d

Description: Law for double surreal subtraction. (Contributed by Scott Fenton, 9-Mar-2025)

Ref Expression
Hypotheses subsubs4d.1 ⊢ ( 𝜑 → 𝐴 ∈ No )
subsubs4d.2 ⊢ ( 𝜑 → 𝐵 ∈ No )
subsubs4d.3 ⊢ ( 𝜑 → 𝐶 ∈ No )
Assertion subsubs4d ( 𝜑 → ( ( 𝐴 -s 𝐵 ) -s 𝐶 ) = ( 𝐴 -s ( 𝐵 +s 𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 subsubs4d.1 ⊢ ( 𝜑 → 𝐴 ∈ No )
2 subsubs4d.2 ⊢ ( 𝜑 → 𝐵 ∈ No )
3 subsubs4d.3 ⊢ ( 𝜑 → 𝐶 ∈ No )
4 2 negscld ⊢ ( 𝜑 → ( -us ‘ 𝐵 ) ∈ No )
5 3 negscld ⊢ ( 𝜑 → ( -us ‘ 𝐶 ) ∈ No )
6 1 4 5 addsassd ⊢ ( 𝜑 → ( ( 𝐴 +s ( -us ‘ 𝐵 ) ) +s ( -us ‘ 𝐶 ) ) = ( 𝐴 +s ( ( -us ‘ 𝐵 ) +s ( -us ‘ 𝐶 ) ) ) )
7 1 2 subsvald ⊢ ( 𝜑 → ( 𝐴 -s 𝐵 ) = ( 𝐴 +s ( -us ‘ 𝐵 ) ) )
8 7 oveq1d ⊢ ( 𝜑 → ( ( 𝐴 -s 𝐵 ) -s 𝐶 ) = ( ( 𝐴 +s ( -us ‘ 𝐵 ) ) -s 𝐶 ) )
9 1 4 addscld ⊢ ( 𝜑 → ( 𝐴 +s ( -us ‘ 𝐵 ) ) ∈ No )
10 9 3 subsvald ⊢ ( 𝜑 → ( ( 𝐴 +s ( -us ‘ 𝐵 ) ) -s 𝐶 ) = ( ( 𝐴 +s ( -us ‘ 𝐵 ) ) +s ( -us ‘ 𝐶 ) ) )
11 8 10 eqtrd ⊢ ( 𝜑 → ( ( 𝐴 -s 𝐵 ) -s 𝐶 ) = ( ( 𝐴 +s ( -us ‘ 𝐵 ) ) +s ( -us ‘ 𝐶 ) ) )
12 2 3 addscld ⊢ ( 𝜑 → ( 𝐵 +s 𝐶 ) ∈ No )
13 1 12 subsvald ⊢ ( 𝜑 → ( 𝐴 -s ( 𝐵 +s 𝐶 ) ) = ( 𝐴 +s ( -us ‘ ( 𝐵 +s 𝐶 ) ) ) )
14 negsdi ⊢ ( ( 𝐵 ∈ No ∧ 𝐶 ∈ No ) → ( -us ‘ ( 𝐵 +s 𝐶 ) ) = ( ( -us ‘ 𝐵 ) +s ( -us ‘ 𝐶 ) ) )
15 2 3 14 syl2anc ⊢ ( 𝜑 → ( -us ‘ ( 𝐵 +s 𝐶 ) ) = ( ( -us ‘ 𝐵 ) +s ( -us ‘ 𝐶 ) ) )
16 15 oveq2d ⊢ ( 𝜑 → ( 𝐴 +s ( -us ‘ ( 𝐵 +s 𝐶 ) ) ) = ( 𝐴 +s ( ( -us ‘ 𝐵 ) +s ( -us ‘ 𝐶 ) ) ) )
17 13 16 eqtrd ⊢ ( 𝜑 → ( 𝐴 -s ( 𝐵 +s 𝐶 ) ) = ( 𝐴 +s ( ( -us ‘ 𝐵 ) +s ( -us ‘ 𝐶 ) ) ) )
18 6 11 17 3eqtr4d ⊢ ( 𝜑 → ( ( 𝐴 -s 𝐵 ) -s 𝐶 ) = ( 𝐴 -s ( 𝐵 +s 𝐶 ) ) )