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Theorem ifeq3da 30898
Description: Given an expression 𝐶 containing if(𝜓, 𝐸, 𝐹), substitute (hypotheses .1 and .2) and evaluate (hypotheses .3 and .4) it for both cases at the same time. (Contributed by Thierry Arnoux, 13-Dec-2021.)
Hypotheses
Ref Expression
ifeq3da.1 (if(𝜓, 𝐸, 𝐹) = 𝐸𝐶 = 𝐺)
ifeq3da.2 (if(𝜓, 𝐸, 𝐹) = 𝐹𝐶 = 𝐻)
ifeq3da.3 (𝜑𝐺 = 𝐴)
ifeq3da.4 (𝜑𝐻 = 𝐵)
Assertion
Ref Expression
ifeq3da (𝜑 → if(𝜓, 𝐴, 𝐵) = 𝐶)

Proof of Theorem ifeq3da
StepHypRef Expression
1 iftrue 4471 . . . . 5 (𝜓 → if(𝜓, 𝐸, 𝐹) = 𝐸)
2 ifeq3da.1 . . . . 5 (if(𝜓, 𝐸, 𝐹) = 𝐸𝐶 = 𝐺)
31, 2syl 17 . . . 4 (𝜓𝐶 = 𝐺)
43adantl 482 . . 3 ((𝜑𝜓) → 𝐶 = 𝐺)
5 ifeq3da.3 . . . 4 (𝜑𝐺 = 𝐴)
65adantr 481 . . 3 ((𝜑𝜓) → 𝐺 = 𝐴)
74, 6eqtr2d 2781 . 2 ((𝜑𝜓) → 𝐴 = 𝐶)
8 iffalse 4474 . . . . 5 𝜓 → if(𝜓, 𝐸, 𝐹) = 𝐹)
9 ifeq3da.2 . . . . 5 (if(𝜓, 𝐸, 𝐹) = 𝐹𝐶 = 𝐻)
108, 9syl 17 . . . 4 𝜓𝐶 = 𝐻)
1110adantl 482 . . 3 ((𝜑 ∧ ¬ 𝜓) → 𝐶 = 𝐻)
12 ifeq3da.4 . . . 4 (𝜑𝐻 = 𝐵)
1312adantr 481 . . 3 ((𝜑 ∧ ¬ 𝜓) → 𝐻 = 𝐵)
1411, 13eqtr2d 2781 . 2 ((𝜑 ∧ ¬ 𝜓) → 𝐵 = 𝐶)
157, 14ifeqda 4501 1 (𝜑 → if(𝜓, 𝐴, 𝐵) = 𝐶)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 396   = wceq 1542  ifcif 4465
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1975  ax-7 2015  ax-8 2112  ax-9 2120  ax-ext 2711
This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-ex 1787  df-sb 2072  df-clab 2718  df-cleq 2732  df-clel 2818  df-if 4466
This theorem is referenced by:  circlemeth  32629
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