Theorem ifsb 4499

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 4491	. . . 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 4491	. . 3 ⊢ (𝜑 → if(𝜑, 𝐷, 𝐸) = 𝐷)
5	3, 4	eqtr4d 2801	. 2 ⊢ (𝜑 → 𝐶 = if(𝜑, 𝐷, 𝐸))
6		iffalse 4494	. . . 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 4494	. . 3 ⊢ (¬ 𝜑 → if(𝜑, 𝐷, 𝐸) = 𝐸)
10	8, 9	eqtr4d 2801	. 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 4485

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-if 4486

This theorem is used by:  fvif  6924  iffv  6925  ovif  7541  ovif2  7542  ifov  7544  xmulneg1  13426  efrlim  27327  lgsneg  27678  lgsdilem  27681  rpvmasum2  27869