Metamath Proof Explorer


Theorem negsubsdi2d

Description: Distribution of negative over subtraction. (Contributed by Scott Fenton, 5-Feb-2025)

Ref Expression
Hypotheses negsubsdi2d.1 ⊢ ( 𝜑 → 𝐴 ∈ No )
negsubsdi2d.2 ⊢ ( 𝜑 → 𝐵 ∈ No )
Assertion negsubsdi2d ( 𝜑 → ( -us ‘ ( 𝐴 -s 𝐵 ) ) = ( 𝐵 -s 𝐴 ) )

Proof

Step Hyp Ref Expression
1 negsubsdi2d.1 ⊢ ( 𝜑 → 𝐴 ∈ No )
2 negsubsdi2d.2 ⊢ ( 𝜑 → 𝐵 ∈ No )
3 2 negscld ⊢ ( 𝜑 → ( -us ‘ 𝐵 ) ∈ No )
4 negsdi ⊢ ( ( 𝐴 ∈ No ∧ ( -us ‘ 𝐵 ) ∈ No ) → ( -us ‘ ( 𝐴 +s ( -us ‘ 𝐵 ) ) ) = ( ( -us ‘ 𝐴 ) +s ( -us ‘ ( -us ‘ 𝐵 ) ) ) )
5 1 3 4 syl2anc ⊢ ( 𝜑 → ( -us ‘ ( 𝐴 +s ( -us ‘ 𝐵 ) ) ) = ( ( -us ‘ 𝐴 ) +s ( -us ‘ ( -us ‘ 𝐵 ) ) ) )
6 negnegs ⊢ ( 𝐵 ∈ No → ( -us ‘ ( -us ‘ 𝐵 ) ) = 𝐵 )
7 2 6 syl ⊢ ( 𝜑 → ( -us ‘ ( -us ‘ 𝐵 ) ) = 𝐵 )
8 7 oveq2d ⊢ ( 𝜑 → ( ( -us ‘ 𝐴 ) +s ( -us ‘ ( -us ‘ 𝐵 ) ) ) = ( ( -us ‘ 𝐴 ) +s 𝐵 ) )
9 1 negscld ⊢ ( 𝜑 → ( -us ‘ 𝐴 ) ∈ No )
10 9 2 addscomd ⊢ ( 𝜑 → ( ( -us ‘ 𝐴 ) +s 𝐵 ) = ( 𝐵 +s ( -us ‘ 𝐴 ) ) )
11 5 8 10 3eqtrd ⊢ ( 𝜑 → ( -us ‘ ( 𝐴 +s ( -us ‘ 𝐵 ) ) ) = ( 𝐵 +s ( -us ‘ 𝐴 ) ) )
12 1 2 subsvald ⊢ ( 𝜑 → ( 𝐴 -s 𝐵 ) = ( 𝐴 +s ( -us ‘ 𝐵 ) ) )
13 12 fveq2d ⊢ ( 𝜑 → ( -us ‘ ( 𝐴 -s 𝐵 ) ) = ( -us ‘ ( 𝐴 +s ( -us ‘ 𝐵 ) ) ) )
14 2 1 subsvald ⊢ ( 𝜑 → ( 𝐵 -s 𝐴 ) = ( 𝐵 +s ( -us ‘ 𝐴 ) ) )
15 11 13 14 3eqtr4d ⊢ ( 𝜑 → ( -us ‘ ( 𝐴 -s 𝐵 ) ) = ( 𝐵 -s 𝐴 ) )