Theorem ififcom 32638

Description: Commute two nested conditionals. (Contributed by Thierry Arnoux, 4-May-2026.)

Assertion

Ref	Expression
ififcom	⊢ if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = if(𝜓, if(𝜑, 𝐴, 𝐵), 𝐵)

Proof of Theorem ififcom

Step	Hyp	Ref	Expression
1		ancom 461	. . 3 ⊢ ((𝜑 ∧ 𝜓) ↔ (𝜓 ∧ 𝜑))
2		ifbi 4477	. . 3 ⊢ (((𝜑 ∧ 𝜓) ↔ (𝜓 ∧ 𝜑)) → if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if((𝜓 ∧ 𝜑), 𝐴, 𝐵))
3	1, 2	ax-mp 5	. 2 ⊢ if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if((𝜓 ∧ 𝜑), 𝐴, 𝐵)
4		ifan 4508	. 2 ⊢ if((𝜑 ∧ 𝜓), 𝐴, 𝐵) = if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵)
5		ifan 4508	. 2 ⊢ if((𝜓 ∧ 𝜑), 𝐴, 𝐵) = if(𝜓, if(𝜑, 𝐴, 𝐵), 𝐵)
6	3, 4, 5	3eqtr3i 2770	1 ⊢ if(𝜑, if(𝜓, 𝐴, 𝐵), 𝐵) = if(𝜓, if(𝜑, 𝐴, 𝐵), 𝐵)

Syntax hints:   ↔ wb 207   ∧ wa 396   = wceq 1547  ifcif 4454

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 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2711

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-ex 1787  df-sb 2074  df-clab 2718  df-cleq 2731  df-clel 2814  df-if 4455

This theorem is referenced by:  mplasclco  33700