Theorem undif1ss 3599

Description: Absorption of difference by union. In classical logic, as Theorem 35 of [Suppes] p. 29, this would be equality rather than subset. (Contributed by Jim Kingdon, 4-Aug-2018.)

Assertion

Ref	Expression
undif1ss	⊢ ((𝐴 ∖ 𝐵) ∪ 𝐵) ⊆ (𝐴 ∪ 𝐵)

Proof of Theorem undif1ss

Step	Hyp	Ref	Expression
1		difss 3355	. 2 ⊢ (𝐴 ∖ 𝐵) ⊆ 𝐴
2		unss1 3398	. 2 ⊢ ((𝐴 ∖ 𝐵) ⊆ 𝐴 → ((𝐴 ∖ 𝐵) ∪ 𝐵) ⊆ (𝐴 ∪ 𝐵))
3	1, 2	ax-mp 5	1 ⊢ ((𝐴 ∖ 𝐵) ∪ 𝐵) ⊆ (𝐴 ∪ 𝐵)

Syntax hints:   ∖ cdif 3217   ∪ cun 3218   ⊆ wss 3220

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-in1 623  ax-in2 624  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This theorem depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823  df-dif 3222  df-un 3224  df-in 3226  df-ss 3233

This theorem is referenced by:  undif2ss  3600  pwundifss  4425