Mathbox for Thierry Arnoux |
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Mirrors > Home > MPE Home > Th. List > Mathboxes > ifeq3da | Structured version Visualization version GIF version |
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.) |
Ref | Expression |
---|---|
ifeq3da.1 | ⊢ (if(𝜓, 𝐸, 𝐹) = 𝐸 → 𝐶 = 𝐺) |
ifeq3da.2 | ⊢ (if(𝜓, 𝐸, 𝐹) = 𝐹 → 𝐶 = 𝐻) |
ifeq3da.3 | ⊢ (𝜑 → 𝐺 = 𝐴) |
ifeq3da.4 | ⊢ (𝜑 → 𝐻 = 𝐵) |
Ref | Expression |
---|---|
ifeq3da | ⊢ (𝜑 → if(𝜓, 𝐴, 𝐵) = 𝐶) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | iftrue 4471 | . . . . 5 ⊢ (𝜓 → if(𝜓, 𝐸, 𝐹) = 𝐸) | |
2 | ifeq3da.1 | . . . . 5 ⊢ (if(𝜓, 𝐸, 𝐹) = 𝐸 → 𝐶 = 𝐺) | |
3 | 1, 2 | syl 17 | . . . 4 ⊢ (𝜓 → 𝐶 = 𝐺) |
4 | 3 | adantl 482 | . . 3 ⊢ ((𝜑 ∧ 𝜓) → 𝐶 = 𝐺) |
5 | ifeq3da.3 | . . . 4 ⊢ (𝜑 → 𝐺 = 𝐴) | |
6 | 5 | adantr 481 | . . 3 ⊢ ((𝜑 ∧ 𝜓) → 𝐺 = 𝐴) |
7 | 4, 6 | eqtr2d 2781 | . 2 ⊢ ((𝜑 ∧ 𝜓) → 𝐴 = 𝐶) |
8 | iffalse 4474 | . . . . 5 ⊢ (¬ 𝜓 → if(𝜓, 𝐸, 𝐹) = 𝐹) | |
9 | ifeq3da.2 | . . . . 5 ⊢ (if(𝜓, 𝐸, 𝐹) = 𝐹 → 𝐶 = 𝐻) | |
10 | 8, 9 | syl 17 | . . . 4 ⊢ (¬ 𝜓 → 𝐶 = 𝐻) |
11 | 10 | adantl 482 | . . 3 ⊢ ((𝜑 ∧ ¬ 𝜓) → 𝐶 = 𝐻) |
12 | ifeq3da.4 | . . . 4 ⊢ (𝜑 → 𝐻 = 𝐵) | |
13 | 12 | adantr 481 | . . 3 ⊢ ((𝜑 ∧ ¬ 𝜓) → 𝐻 = 𝐵) |
14 | 11, 13 | eqtr2d 2781 | . 2 ⊢ ((𝜑 ∧ ¬ 𝜓) → 𝐵 = 𝐶) |
15 | 7, 14 | ifeqda 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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