Metamath Proof Explorer
Description: Deduction associated with suceq . (Contributed by Rohan Ridenour, 8-Aug-2023)
|
|
Ref |
Expression |
|
Hypothesis |
suceqd.1 |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
|
Assertion |
suceqd |
⊢ ( 𝜑 → suc 𝐴 = suc 𝐵 ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
suceqd.1 |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
| 2 |
1
|
sneqd |
⊢ ( 𝜑 → { 𝐴 } = { 𝐵 } ) |
| 3 |
1 2
|
uneq12d |
⊢ ( 𝜑 → ( 𝐴 ∪ { 𝐴 } ) = ( 𝐵 ∪ { 𝐵 } ) ) |
| 4 |
|
df-suc |
⊢ suc 𝐴 = ( 𝐴 ∪ { 𝐴 } ) |
| 5 |
|
df-suc |
⊢ suc 𝐵 = ( 𝐵 ∪ { 𝐵 } ) |
| 6 |
3 4 5
|
3eqtr4g |
⊢ ( 𝜑 → suc 𝐴 = suc 𝐵 ) |