Theorem ifidss 3541

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 3540	. 2 ⊢ if(𝜑, 𝐴, 𝐴) ⊆ (𝐴 ∪ 𝐴)
2		unidm 3270	. 2 ⊢ (𝐴 ∪ 𝐴) = 𝐴
3	1, 2	sseqtri 3181	1 ⊢ if(𝜑, 𝐴, 𝐴) ⊆ 𝐴

Syntax hints:   ∪ cun 3119   ⊆ wss 3121  ifcif 3526

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 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-rab 2457  df-v 2732  df-un 3125  df-in 3127  df-ss 3134  df-if 3527

This theorem is referenced by: (None)