Metamath Proof Explorer


Theorem reltsub1

Description: Subtraction from both sides of 'less than'. Compare ltsub1 . (Contributed by SN, 13-Feb-2024)

Ref Expression
Assertion reltsub1 ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐴 < 𝐵 ↔ ( 𝐴 𝐶 ) < ( 𝐵 𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 rersubcl ( ( 𝐴 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐴 𝐶 ) ∈ ℝ )
2 1 3adant2 ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐴 𝐶 ) ∈ ℝ )
3 rersubcl ( ( 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐵 𝐶 ) ∈ ℝ )
4 3 3adant1 ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐵 𝐶 ) ∈ ℝ )
5 simp3 ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → 𝐶 ∈ ℝ )
6 2 4 5 ltadd2d ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ( 𝐴 𝐶 ) < ( 𝐵 𝐶 ) ↔ ( 𝐶 + ( 𝐴 𝐶 ) ) < ( 𝐶 + ( 𝐵 𝐶 ) ) ) )
7 repncan3 ( ( 𝐶 ∈ ℝ ∧ 𝐴 ∈ ℝ ) → ( 𝐶 + ( 𝐴 𝐶 ) ) = 𝐴 )
8 7 ancoms ( ( 𝐴 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐶 + ( 𝐴 𝐶 ) ) = 𝐴 )
9 8 3adant2 ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐶 + ( 𝐴 𝐶 ) ) = 𝐴 )
10 repncan3 ( ( 𝐶 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( 𝐶 + ( 𝐵 𝐶 ) ) = 𝐵 )
11 10 ancoms ( ( 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐶 + ( 𝐵 𝐶 ) ) = 𝐵 )
12 11 3adant1 ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐶 + ( 𝐵 𝐶 ) ) = 𝐵 )
13 9 12 breq12d ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ( 𝐶 + ( 𝐴 𝐶 ) ) < ( 𝐶 + ( 𝐵 𝐶 ) ) ↔ 𝐴 < 𝐵 ) )
14 6 13 bitr2d ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐴 < 𝐵 ↔ ( 𝐴 𝐶 ) < ( 𝐵 𝐶 ) ) )