Theorem ifsb 4495

Description: Distribute a function over an if-clause. (Contributed by Mario Carneiro, 14-Aug-2013.)

Hypotheses

Ref	Expression
ifsb.1	⊢ (if(𝜑, 𝐴, 𝐵) = 𝐴 → 𝐶 = 𝐷)
ifsb.2	⊢ (if(𝜑, 𝐴, 𝐵) = 𝐵 → 𝐶 = 𝐸)

Assertion

Ref	Expression
ifsb	⊢ 𝐶 = if(𝜑, 𝐷, 𝐸)

Proof of Theorem ifsb

Step	Hyp	Ref	Expression
1		iftrue 4487	. . . 4 ⊢ (𝜑 → if(𝜑, 𝐴, 𝐵) = 𝐴)
2		ifsb.1	. . . 4 ⊢ (if(𝜑, 𝐴, 𝐵) = 𝐴 → 𝐶 = 𝐷)
3	1, 2	syl 18	. . 3 ⊢ (𝜑 → 𝐶 = 𝐷)
4		iftrue 4487	. . 3 ⊢ (𝜑 → if(𝜑, 𝐷, 𝐸) = 𝐷)
5	3, 4	eqtr4d 2798	. 2 ⊢ (𝜑 → 𝐶 = if(𝜑, 𝐷, 𝐸))
6		iffalse 4490	. . . 4 ⊢ (¬ 𝜑 → if(𝜑, 𝐴, 𝐵) = 𝐵)
7		ifsb.2	. . . 4 ⊢ (if(𝜑, 𝐴, 𝐵) = 𝐵 → 𝐶 = 𝐸)
8	6, 7	syl 18	. . 3 ⊢ (¬ 𝜑 → 𝐶 = 𝐸)
9		iffalse 4490	. . 3 ⊢ (¬ 𝜑 → if(𝜑, 𝐷, 𝐸) = 𝐸)
10	8, 9	eqtr4d 2798	. 2 ⊢ (¬ 𝜑 → 𝐶 = if(𝜑, 𝐷, 𝐸))
11	5, 10	pm2.61i 184	1 ⊢ 𝐶 = if(𝜑, 𝐷, 𝐸)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   = wceq 1570  ifcif 4481

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-if 4482

This theorem is used by:  fvif  6889  iffv  6890  ovif  7506  ovif2  7507  ifov  7509  xmulneg1  13368  efrlim  27260  lgsneg  27611  lgsdilem  27614  rpvmasum2  27802