Theorem ifidss 3521

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 3520	. 2 ⊢ if(𝜑, 𝐴, 𝐴) ⊆ (𝐴 ∪ 𝐴)
2		unidm 3251	. 2 ⊢ (𝐴 ∪ 𝐴) = 𝐴
3	1, 2	sseqtri 3162	1 ⊢ if(𝜑, 𝐴, 𝐴) ⊆ 𝐴

Syntax hints:   ∪ cun 3100   ⊆ wss 3102  ifcif 3506

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1427  ax-7 1428  ax-gen 1429  ax-ie1 1473  ax-ie2 1474  ax-8 1484  ax-10 1485  ax-11 1486  ax-i12 1487  ax-bndl 1489  ax-4 1490  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2139

This theorem depends on definitions:  df-bi 116  df-tru 1338  df-nf 1441  df-sb 1743  df-clab 2144  df-cleq 2150  df-clel 2153  df-nfc 2288  df-rab 2444  df-v 2714  df-un 3106  df-in 3108  df-ss 3115  df-if 3507

This theorem is referenced by: (None)