Theorem ifidss 3625

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 ( ph , A , A ) C_ A

Proof of Theorem ifidss

Step	Hyp	Ref	Expression
1		ifssun 3624	. 2 |- if ( ph , A , A ) C_ ( A u. A )
2		unidm 3352	. 2 |- ( A u. A ) = A
3	1, 2	sseqtri 3262	1 |- if ( ph , A , A ) C_ A

Syntax hints:   u. cun 3199   C_ wss 3201   ifcif 3607

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 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2364  df-rab 2520  df-v 2805  df-un 3205  df-in 3207  df-ss 3214  df-if 3608

This theorem is referenced by: (None)