Metamath Proof Explorer
Description: Deduction doubly substituting both sides of a biconditional.
(Contributed by AV, 30-Jul-2023)
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Ref |
Expression |
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Hypotheses |
sbbid.1 |
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sbbid.2 |
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2sbbid.1 |
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Assertion |
2sbbid |
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Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
sbbid.1 |
|
2 |
|
sbbid.2 |
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3 |
|
2sbbid.1 |
|
4 |
3 2
|
sbbid |
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5 |
1 4
|
sbbid |
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