Theorem jaeqifi 35710

Description: Inference combining four equality antecedents into one equality of conditional operators antecedent. (Contributed by BTernaryTau, 14-Jul-2026.)

Hypotheses

Ref	Expression
jaeqifi.1	⊢ (𝐴 = 𝐶 → 𝜑)
jaeqifi.2	⊢ (𝐴 = 𝐷 → 𝜑)
jaeqifi.3	⊢ (𝐵 = 𝐶 → 𝜑)
jaeqifi.4	⊢ (𝐵 = 𝐷 → 𝜑)

Assertion

Ref	Expression
jaeqifi	⊢ (if(𝜓, 𝐴, 𝐵) = if(𝜒, 𝐶, 𝐷) → 𝜑)

Proof of Theorem jaeqifi

Step	Hyp	Ref	Expression
1		iftrue 4488	. . . 4 ⊢ (𝜓 → if(𝜓, 𝐴, 𝐵) = 𝐴)
2		iftrue 4488	. . . 4 ⊢ (𝜒 → if(𝜒, 𝐶, 𝐷) = 𝐶)
3	1, 2	eqeqan12d 2775	. . 3 ⊢ ((𝜓 ∧ 𝜒) → (if(𝜓, 𝐴, 𝐵) = if(𝜒, 𝐶, 𝐷) ↔ 𝐴 = 𝐶))
4		jaeqifi.1	. . 3 ⊢ (𝐴 = 𝐶 → 𝜑)
5	3, 4	biimtrdi 256	. 2 ⊢ ((𝜓 ∧ 𝜒) → (if(𝜓, 𝐴, 𝐵) = if(𝜒, 𝐶, 𝐷) → 𝜑))
6		iffalse 4491	. . . 4 ⊢ (¬ 𝜒 → if(𝜒, 𝐶, 𝐷) = 𝐷)
7	1, 6	eqeqan12d 2775	. . 3 ⊢ ((𝜓 ∧ ¬ 𝜒) → (if(𝜓, 𝐴, 𝐵) = if(𝜒, 𝐶, 𝐷) ↔ 𝐴 = 𝐷))
8		jaeqifi.2	. . 3 ⊢ (𝐴 = 𝐷 → 𝜑)
9	7, 8	biimtrdi 256	. 2 ⊢ ((𝜓 ∧ ¬ 𝜒) → (if(𝜓, 𝐴, 𝐵) = if(𝜒, 𝐶, 𝐷) → 𝜑))
10		iffalse 4491	. . . 4 ⊢ (¬ 𝜓 → if(𝜓, 𝐴, 𝐵) = 𝐵)
11	10, 2	eqeqan12d 2775	. . 3 ⊢ ((¬ 𝜓 ∧ 𝜒) → (if(𝜓, 𝐴, 𝐵) = if(𝜒, 𝐶, 𝐷) ↔ 𝐵 = 𝐶))
12		jaeqifi.3	. . 3 ⊢ (𝐵 = 𝐶 → 𝜑)
13	11, 12	biimtrdi 256	. 2 ⊢ ((¬ 𝜓 ∧ 𝜒) → (if(𝜓, 𝐴, 𝐵) = if(𝜒, 𝐶, 𝐷) → 𝜑))
14	10, 6	eqeqan12d 2775	. . 3 ⊢ ((¬ 𝜓 ∧ ¬ 𝜒) → (if(𝜓, 𝐴, 𝐵) = if(𝜒, 𝐶, 𝐷) ↔ 𝐵 = 𝐷))
15		jaeqifi.4	. . 3 ⊢ (𝐵 = 𝐷 → 𝜑)
16	14, 15	biimtrdi 256	. 2 ⊢ ((¬ 𝜓 ∧ ¬ 𝜒) → (if(𝜓, 𝐴, 𝐵) = if(𝜒, 𝐶, 𝐷) → 𝜑))
17	5, 9, 13, 16	4cases 1056	1 ⊢ (if(𝜓, 𝐴, 𝐵) = if(𝜒, 𝐶, 𝐷) → 𝜑)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   = wceq 1570  ifcif 4482

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 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-if 4483

This theorem is used by:  soinfdom  35717