Metamath Proof Explorer


Theorem addlsub

Description: Left-subtraction: Subtraction of the left summand from the result of an addition. (Contributed by BJ, 6-Jun-2019)

Ref Expression
Hypotheses addlsub.a ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
addlsub.b ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
addlsub.c ⊢ ( 𝜑 → 𝐶 ∈ ℂ )
Assertion addlsub ( 𝜑 → ( ( 𝐴 + 𝐵 ) = 𝐶 ↔ 𝐴 = ( 𝐶 − 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 addlsub.a ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
2 addlsub.b ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
3 addlsub.c ⊢ ( 𝜑 → 𝐶 ∈ ℂ )
4 oveq1 ⊢ ( ( 𝐴 + 𝐵 ) = 𝐶 → ( ( 𝐴 + 𝐵 ) − 𝐵 ) = ( 𝐶 − 𝐵 ) )
5 1 2 pncand ⊢ ( 𝜑 → ( ( 𝐴 + 𝐵 ) − 𝐵 ) = 𝐴 )
6 eqtr2 ⊢ ( ( ( ( 𝐴 + 𝐵 ) − 𝐵 ) = ( 𝐶 − 𝐵 ) ∧ ( ( 𝐴 + 𝐵 ) − 𝐵 ) = 𝐴 ) → ( 𝐶 − 𝐵 ) = 𝐴 )
7 6 eqcomd ⊢ ( ( ( ( 𝐴 + 𝐵 ) − 𝐵 ) = ( 𝐶 − 𝐵 ) ∧ ( ( 𝐴 + 𝐵 ) − 𝐵 ) = 𝐴 ) → 𝐴 = ( 𝐶 − 𝐵 ) )
8 7 a1i ⊢ ( 𝜑 → ( ( ( ( 𝐴 + 𝐵 ) − 𝐵 ) = ( 𝐶 − 𝐵 ) ∧ ( ( 𝐴 + 𝐵 ) − 𝐵 ) = 𝐴 ) → 𝐴 = ( 𝐶 − 𝐵 ) ) )
9 5 8 mpan2d ⊢ ( 𝜑 → ( ( ( 𝐴 + 𝐵 ) − 𝐵 ) = ( 𝐶 − 𝐵 ) → 𝐴 = ( 𝐶 − 𝐵 ) ) )
10 4 9 syl5 ⊢ ( 𝜑 → ( ( 𝐴 + 𝐵 ) = 𝐶 → 𝐴 = ( 𝐶 − 𝐵 ) ) )
11 oveq1 ⊢ ( 𝐴 = ( 𝐶 − 𝐵 ) → ( 𝐴 + 𝐵 ) = ( ( 𝐶 − 𝐵 ) + 𝐵 ) )
12 3 2 npcand ⊢ ( 𝜑 → ( ( 𝐶 − 𝐵 ) + 𝐵 ) = 𝐶 )
13 eqtr ⊢ ( ( ( 𝐴 + 𝐵 ) = ( ( 𝐶 − 𝐵 ) + 𝐵 ) ∧ ( ( 𝐶 − 𝐵 ) + 𝐵 ) = 𝐶 ) → ( 𝐴 + 𝐵 ) = 𝐶 )
14 13 a1i ⊢ ( 𝜑 → ( ( ( 𝐴 + 𝐵 ) = ( ( 𝐶 − 𝐵 ) + 𝐵 ) ∧ ( ( 𝐶 − 𝐵 ) + 𝐵 ) = 𝐶 ) → ( 𝐴 + 𝐵 ) = 𝐶 ) )
15 12 14 mpan2d ⊢ ( 𝜑 → ( ( 𝐴 + 𝐵 ) = ( ( 𝐶 − 𝐵 ) + 𝐵 ) → ( 𝐴 + 𝐵 ) = 𝐶 ) )
16 11 15 syl5 ⊢ ( 𝜑 → ( 𝐴 = ( 𝐶 − 𝐵 ) → ( 𝐴 + 𝐵 ) = 𝐶 ) )
17 10 16 impbid ⊢ ( 𝜑 → ( ( 𝐴 + 𝐵 ) = 𝐶 ↔ 𝐴 = ( 𝐶 − 𝐵 ) ) )