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Metamath Proof Explorer
Description: Substitution of equality into both sides of an inequality. (Contributed by NM, 24-Jul-2012) (Proof shortened by Wolf Lammen, 21-Nov-2019)
|
|
Ref |
Expression |
|
Hypotheses |
3netr4d.1 |
|- ( ph -> A =/= B ) |
|
|
3netr4d.2 |
|- ( ph -> C = A ) |
|
|
3netr4d.3 |
|- ( ph -> D = B ) |
|
Assertion |
3netr4d |
|- ( ph -> C =/= D ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
3netr4d.1 |
|- ( ph -> A =/= B ) |
| 2 |
|
3netr4d.2 |
|- ( ph -> C = A ) |
| 3 |
|
3netr4d.3 |
|- ( ph -> D = B ) |
| 4 |
2 1
|
eqnetrd |
|- ( ph -> C =/= B ) |
| 5 |
4 3
|
neeqtrrd |
|- ( ph -> C =/= D ) |