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Description: Law for addition and subtraction. (Contributed by Steven Nguyen, 28-Jan-2023)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | readdsub | ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ( 𝐴 + 𝐵 ) −ℝ 𝐶 ) = ( ( 𝐴 −ℝ 𝐶 ) + 𝐵 ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | simp3 | ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → 𝐶 ∈ ℝ ) | |
| 2 | readdcl | ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( 𝐴 + 𝐵 ) ∈ ℝ ) | |
| 3 | 2 | 3adant3 | ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐴 + 𝐵 ) ∈ ℝ ) |
| 4 | repncan3 | ⊢ ( ( 𝐶 ∈ ℝ ∧ ( 𝐴 + 𝐵 ) ∈ ℝ ) → ( 𝐶 + ( ( 𝐴 + 𝐵 ) −ℝ 𝐶 ) ) = ( 𝐴 + 𝐵 ) ) | |
| 5 | 1 3 4 | syl2anc | ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐶 + ( ( 𝐴 + 𝐵 ) −ℝ 𝐶 ) ) = ( 𝐴 + 𝐵 ) ) |
| 6 | repncan3 | ⊢ ( ( 𝐶 ∈ ℝ ∧ 𝐴 ∈ ℝ ) → ( 𝐶 + ( 𝐴 −ℝ 𝐶 ) ) = 𝐴 ) | |
| 7 | 6 | ancoms | ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐶 + ( 𝐴 −ℝ 𝐶 ) ) = 𝐴 ) |
| 8 | 7 | 3adant2 | ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐶 + ( 𝐴 −ℝ 𝐶 ) ) = 𝐴 ) |
| 9 | 8 | oveq1d | ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ( 𝐶 + ( 𝐴 −ℝ 𝐶 ) ) + 𝐵 ) = ( 𝐴 + 𝐵 ) ) |
| 10 | 1 | recnd | ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → 𝐶 ∈ ℂ ) |
| 11 | rersubcl | ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐴 −ℝ 𝐶 ) ∈ ℝ ) | |
| 12 | 11 | 3adant2 | ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐴 −ℝ 𝐶 ) ∈ ℝ ) |
| 13 | 12 | recnd | ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐴 −ℝ 𝐶 ) ∈ ℂ ) |
| 14 | simp2 | ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → 𝐵 ∈ ℝ ) | |
| 15 | 14 | recnd | ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → 𝐵 ∈ ℂ ) |
| 16 | 10 13 15 | addassd | ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ( 𝐶 + ( 𝐴 −ℝ 𝐶 ) ) + 𝐵 ) = ( 𝐶 + ( ( 𝐴 −ℝ 𝐶 ) + 𝐵 ) ) ) |
| 17 | 5 9 16 | 3eqtr2d | ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐶 + ( ( 𝐴 + 𝐵 ) −ℝ 𝐶 ) ) = ( 𝐶 + ( ( 𝐴 −ℝ 𝐶 ) + 𝐵 ) ) ) |
| 18 | rersubcl | ⊢ ( ( ( 𝐴 + 𝐵 ) ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ( 𝐴 + 𝐵 ) −ℝ 𝐶 ) ∈ ℝ ) | |
| 19 | 3 1 18 | syl2anc | ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ( 𝐴 + 𝐵 ) −ℝ 𝐶 ) ∈ ℝ ) |
| 20 | 12 14 | readdcld | ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ( 𝐴 −ℝ 𝐶 ) + 𝐵 ) ∈ ℝ ) |
| 21 | readdcan | ⊢ ( ( ( ( 𝐴 + 𝐵 ) −ℝ 𝐶 ) ∈ ℝ ∧ ( ( 𝐴 −ℝ 𝐶 ) + 𝐵 ) ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ( 𝐶 + ( ( 𝐴 + 𝐵 ) −ℝ 𝐶 ) ) = ( 𝐶 + ( ( 𝐴 −ℝ 𝐶 ) + 𝐵 ) ) ↔ ( ( 𝐴 + 𝐵 ) −ℝ 𝐶 ) = ( ( 𝐴 −ℝ 𝐶 ) + 𝐵 ) ) ) | |
| 22 | 19 20 1 21 | syl3anc | ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ( 𝐶 + ( ( 𝐴 + 𝐵 ) −ℝ 𝐶 ) ) = ( 𝐶 + ( ( 𝐴 −ℝ 𝐶 ) + 𝐵 ) ) ↔ ( ( 𝐴 + 𝐵 ) −ℝ 𝐶 ) = ( ( 𝐴 −ℝ 𝐶 ) + 𝐵 ) ) ) |
| 23 | 17 22 | mpbid | ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ( 𝐴 + 𝐵 ) −ℝ 𝐶 ) = ( ( 𝐴 −ℝ 𝐶 ) + 𝐵 ) ) |