Theorem undif2ss 3293

Description: Absorption of difference by union. In classical logic, as in Part of proof of Corollary 6K of [Enderton] p. 144, this would be equality rather than subset. (Contributed by Jim Kingdon, 4-Aug-2018.)

Assertion

Ref	Expression
undif2ss	⊢ (A ∪ (B ∖ A)) ⊆ (A ∪ B)

Proof of Theorem undif2ss

Step	Hyp	Ref	Expression
1		undif1ss 3292	. 2 ⊢ ((B ∖ A) ∪ A) ⊆ (B ∪ A)
2		uncom 3081	. 2 ⊢ (A ∪ (B ∖ A)) = ((B ∖ A) ∪ A)
3		uncom 3081	. 2 ⊢ (A ∪ B) = (B ∪ A)
4	1, 2, 3	3sstr4i 2978	1 ⊢ (A ∪ (B ∖ A)) ⊆ (A ∪ B)

Syntax hints:   ∖ cdif 2908   ∪ cun 2909   ⊆ wss 2911

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bndl 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019

This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-v 2553  df-dif 2914  df-un 2916  df-in 2918  df-ss 2925

This theorem is referenced by: (None)