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Description: ((a-b)+c)-a = c-a holds for complex numbers a,b,c. (Contributed by Alexander van der Vekens, 23-Mar-2018)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | cnambpcma | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐴 − 𝐵 ) + 𝐶 ) − 𝐴 ) = ( 𝐶 − 𝐵 ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | subcl | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 − 𝐵 ) ∈ ℂ ) | |
| 2 | 1 | 3adant3 | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 − 𝐵 ) ∈ ℂ ) |
| 3 | simp3 | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐶 ∈ ℂ ) | |
| 4 | simp1 | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐴 ∈ ℂ ) | |
| 5 | 2 3 4 | addsubd | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐴 − 𝐵 ) + 𝐶 ) − 𝐴 ) = ( ( ( 𝐴 − 𝐵 ) − 𝐴 ) + 𝐶 ) ) |
| 6 | simpl | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → 𝐴 ∈ ℂ ) | |
| 7 | simpr | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → 𝐵 ∈ ℂ ) | |
| 8 | 6 7 6 | 3jca | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ) ) |
| 9 | 8 | 3adant3 | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ) ) |
| 10 | sub32 | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) − 𝐴 ) = ( ( 𝐴 − 𝐴 ) − 𝐵 ) ) | |
| 11 | 9 10 | syl | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) − 𝐴 ) = ( ( 𝐴 − 𝐴 ) − 𝐵 ) ) |
| 12 | 11 | oveq1d | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐴 − 𝐵 ) − 𝐴 ) + 𝐶 ) = ( ( ( 𝐴 − 𝐴 ) − 𝐵 ) + 𝐶 ) ) |
| 13 | subcl | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ( 𝐴 − 𝐴 ) ∈ ℂ ) | |
| 14 | 13 | anidms | ⊢ ( 𝐴 ∈ ℂ → ( 𝐴 − 𝐴 ) ∈ ℂ ) |
| 15 | 14 | 3ad2ant1 | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 − 𝐴 ) ∈ ℂ ) |
| 16 | simp2 | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐵 ∈ ℂ ) | |
| 17 | 15 16 3 | subadd23d | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐴 − 𝐴 ) − 𝐵 ) + 𝐶 ) = ( ( 𝐴 − 𝐴 ) + ( 𝐶 − 𝐵 ) ) ) |
| 18 | subid | ⊢ ( 𝐴 ∈ ℂ → ( 𝐴 − 𝐴 ) = 0 ) | |
| 19 | 18 | oveq1d | ⊢ ( 𝐴 ∈ ℂ → ( ( 𝐴 − 𝐴 ) + ( 𝐶 − 𝐵 ) ) = ( 0 + ( 𝐶 − 𝐵 ) ) ) |
| 20 | 19 | 3ad2ant1 | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐴 ) + ( 𝐶 − 𝐵 ) ) = ( 0 + ( 𝐶 − 𝐵 ) ) ) |
| 21 | subcl | ⊢ ( ( 𝐶 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐶 − 𝐵 ) ∈ ℂ ) | |
| 22 | 21 | ancoms | ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐶 − 𝐵 ) ∈ ℂ ) |
| 23 | 22 | addlidd | ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 0 + ( 𝐶 − 𝐵 ) ) = ( 𝐶 − 𝐵 ) ) |
| 24 | 23 | 3adant1 | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 0 + ( 𝐶 − 𝐵 ) ) = ( 𝐶 − 𝐵 ) ) |
| 25 | 17 20 24 | 3eqtrd | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐴 − 𝐴 ) − 𝐵 ) + 𝐶 ) = ( 𝐶 − 𝐵 ) ) |
| 26 | 5 12 25 | 3eqtrd | ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐴 − 𝐵 ) + 𝐶 ) − 𝐴 ) = ( 𝐶 − 𝐵 ) ) |