Metamath Proof Explorer


Theorem subsid1

Description: Identity law for subtraction. (Contributed by Scott Fenton, 3-Feb-2025)

Ref Expression
Assertion subsid1 ( 𝐴 ∈ No → ( 𝐴 -s 0s ) = 𝐴 )

Proof

Step Hyp Ref Expression
1 0no ⊢ 0s ∈ No
2 subsval ⊢ ( ( 𝐴 ∈ No ∧ 0s ∈ No ) → ( 𝐴 -s 0s ) = ( 𝐴 +s ( -us ‘ 0s ) ) )
3 1 2 mpan2 ⊢ ( 𝐴 ∈ No → ( 𝐴 -s 0s ) = ( 𝐴 +s ( -us ‘ 0s ) ) )
4 neg0s ⊢ ( -us ‘ 0s ) = 0s
5 4 oveq2i ⊢ ( 𝐴 +s ( -us ‘ 0s ) ) = ( 𝐴 +s 0s )
6 addsrid ⊢ ( 𝐴 ∈ No → ( 𝐴 +s 0s ) = 𝐴 )
7 5 6 eqtrid ⊢ ( 𝐴 ∈ No → ( 𝐴 +s ( -us ‘ 0s ) ) = 𝐴 )
8 3 7 eqtrd ⊢ ( 𝐴 ∈ No → ( 𝐴 -s 0s ) = 𝐴 )