Theorem ifidss 3653

Description: A conditional class whose two alternatives are equal is included in that alternative. With excluded middle, we can prove it is equal to it. (Contributed by BJ, 15-Aug-2024.)

Assertion

Ref	Expression
ifidss	⊢ if(𝜑, 𝐴, 𝐴) ⊆ 𝐴

Proof of Theorem ifidss

Step	Hyp	Ref	Expression
1		ifssun 3652	. 2 ⊢ if(𝜑, 𝐴, 𝐴) ⊆ (𝐴 ∪ 𝐴)
2		unidm 3372	. 2 ⊢ (𝐴 ∪ 𝐴) = 𝐴
3	1, 2	sseqtri 3282	1 ⊢ if(𝜑, 𝐴, 𝐴) ⊆ 𝐴

Syntax hints:   ∪ cun 3218   ⊆ wss 3220  ifcif 3635

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This theorem depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-rab 2537  df-v 2823  df-un 3224  df-in 3226  df-ss 3233  df-if 3636

This theorem is referenced by: (None)