Theorem equncomi 3128

Description: Inference form of equncom 3127. (Contributed by Alan Sare, 18-Feb-2012.)

Hypothesis

Ref	Expression
equncomi.1	⊢ 𝐴 = (𝐵 ∪ 𝐶)

Assertion

Ref	Expression
equncomi	⊢ 𝐴 = (𝐶 ∪ 𝐵)

Proof of Theorem equncomi

Step	Hyp	Ref	Expression
1		equncomi.1	. 2 ⊢ 𝐴 = (𝐵 ∪ 𝐶)
2		equncom 3127	. 2 ⊢ (𝐴 = (𝐵 ∪ 𝐶) ↔ 𝐴 = (𝐶 ∪ 𝐵))
3	1, 2	mpbi 143	1 ⊢ 𝐴 = (𝐶 ∪ 𝐵)

Syntax hints:   = wceq 1285   ∪ cun 2980

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-bndl 1440  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2065

This theorem depends on definitions:  df-bi 115  df-tru 1288  df-nf 1391  df-sb 1688  df-clab 2070  df-cleq 2076  df-clel 2079  df-nfc 2212  df-v 2612  df-un 2986

This theorem is referenced by:  disjssun  3323  difprsn1  3544  unidmrn  4900  phplem1  6409