Theorem ifidss 3587

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 3586	. 2 ⊢ if(𝜑, 𝐴, 𝐴) ⊆ (𝐴 ∪ 𝐴)
2		unidm 3317	. 2 ⊢ (𝐴 ∪ 𝐴) = 𝐴
3	1, 2	sseqtri 3228	1 ⊢ if(𝜑, 𝐴, 𝐴) ⊆ 𝐴

Syntax hints:   ∪ cun 3165   ⊆ wss 3167  ifcif 3572

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 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2188

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-nfc 2338  df-rab 2494  df-v 2775  df-un 3171  df-in 3173  df-ss 3180  df-if 3573

This theorem is referenced by: (None)