Theorem ifidss 3564

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 3563	. 2 ⊢ if(𝜑, 𝐴, 𝐴) ⊆ (𝐴 ∪ 𝐴)
2		unidm 3293	. 2 ⊢ (𝐴 ∪ 𝐴) = 𝐴
3	1, 2	sseqtri 3204	1 ⊢ if(𝜑, 𝐴, 𝐴) ⊆ 𝐴

Syntax hints:   ∪ cun 3142   ⊆ wss 3144  ifcif 3549

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 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2171

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1472  df-sb 1774  df-clab 2176  df-cleq 2182  df-clel 2185  df-nfc 2321  df-rab 2477  df-v 2754  df-un 3148  df-in 3150  df-ss 3157  df-if 3550

This theorem is referenced by: (None)