Theorem phiun 4615

Description: The phi operation distributes over union. (Contributed by SF, 20-Feb-2015.)

Assertion

Ref	Expression
phiun	⊢ Phi (A ∪ B) = ( Phi A ∪ Phi B)

Proof of Theorem phiun

Dummy variables x y are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		rexun 3444	. . 3 ⊢ (∃y ∈ (A ∪ B)x = if(y ∈ Nn , (y +c 1c), y) ↔ (∃y ∈ A x = if(y ∈ Nn , (y +c 1c), y) ∨ ∃y ∈ B x = if(y ∈ Nn , (y +c 1c), y)))
2	1	abbii 2466	. 2 ⊢ {x ∣ ∃y ∈ (A ∪ B)x = if(y ∈ Nn , (y +c 1c), y)} = {x ∣ (∃y ∈ A x = if(y ∈ Nn , (y +c 1c), y) ∨ ∃y ∈ B x = if(y ∈ Nn , (y +c 1c), y))}
3		df-phi 4566	. 2 ⊢ Phi (A ∪ B) = {x ∣ ∃y ∈ (A ∪ B)x = if(y ∈ Nn , (y +c 1c), y)}
4		df-phi 4566	. . . 4 ⊢ Phi A = {x ∣ ∃y ∈ A x = if(y ∈ Nn , (y +c 1c), y)}
5		df-phi 4566	. . . 4 ⊢ Phi B = {x ∣ ∃y ∈ B x = if(y ∈ Nn , (y +c 1c), y)}
6	4, 5	uneq12i 3417	. . 3 ⊢ ( Phi A ∪ Phi B) = ({x ∣ ∃y ∈ A x = if(y ∈ Nn , (y +c 1c), y)} ∪ {x ∣ ∃y ∈ B x = if(y ∈ Nn , (y +c 1c), y)})
7		unab 3522	. . 3 ⊢ ({x ∣ ∃y ∈ A x = if(y ∈ Nn , (y +c 1c), y)} ∪ {x ∣ ∃y ∈ B x = if(y ∈ Nn , (y +c 1c), y)}) = {x ∣ (∃y ∈ A x = if(y ∈ Nn , (y +c 1c), y) ∨ ∃y ∈ B x = if(y ∈ Nn , (y +c 1c), y))}
8	6, 7	eqtri 2373	. 2 ⊢ ( Phi A ∪ Phi B) = {x ∣ (∃y ∈ A x = if(y ∈ Nn , (y +c 1c), y) ∨ ∃y ∈ B x = if(y ∈ Nn , (y +c 1c), y))}
9	2, 3, 8	3eqtr4i 2383	1 ⊢ Phi (A ∪ B) = ( Phi A ∪ Phi B)

This proof depends on syntax axioms:   ∨ wo 357   = wceq 1642   ∈ wcel 1710  {cab 2339  ∃wrex 2616   ∪ cun 3208   ifcif 3663  1_cc1c 4135   Nn cnnc 4374   +_c cplc 4376   Phi cphi 4563

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-rex 2621  df-v 2862  df-nin 3212  df-compl 3213  df-un 3215  df-phi 4566

This theorem is used by:  phialllem2  4618