Metamath Proof Explorer
Description: Equality of successors. (Contributed by NM, 30-Aug-1993) (Proof
shortened by Andrew Salmon, 25-Jul-2011)
|
|
Ref |
Expression |
|
Assertion |
suceq |
⊢ ( 𝐴 = 𝐵 → suc 𝐴 = suc 𝐵 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
id |
⊢ ( 𝐴 = 𝐵 → 𝐴 = 𝐵 ) |
2 |
|
sneq |
⊢ ( 𝐴 = 𝐵 → { 𝐴 } = { 𝐵 } ) |
3 |
1 2
|
uneq12d |
⊢ ( 𝐴 = 𝐵 → ( 𝐴 ∪ { 𝐴 } ) = ( 𝐵 ∪ { 𝐵 } ) ) |
4 |
|
df-suc |
⊢ suc 𝐴 = ( 𝐴 ∪ { 𝐴 } ) |
5 |
|
df-suc |
⊢ suc 𝐵 = ( 𝐵 ∪ { 𝐵 } ) |
6 |
3 4 5
|
3eqtr4g |
⊢ ( 𝐴 = 𝐵 → suc 𝐴 = suc 𝐵 ) |