Theorem elimif 3692

Description: Elimination of a conditional operator contained in a wff ψ. (Contributed by NM, 15-Feb-2005.)

Hypotheses

Ref	Expression
elimif.1	⊢ ( if(φ, A, B) = A → (ψ ↔ χ))
elimif.2	⊢ ( if(φ, A, B) = B → (ψ ↔ θ))

Assertion

Ref	Expression
elimif	⊢ (ψ ↔ ((φ ∧ χ) ∨ (¬ φ ∧ θ)))

Proof of Theorem elimif

Step	Hyp	Ref	Expression
1		exmid 404	. . 3 ⊢ (φ ∨ ¬ φ)
2	1	biantrur 492	. 2 ⊢ (ψ ↔ ((φ ∨ ¬ φ) ∧ ψ))
3		andir 838	. 2 ⊢ (((φ ∨ ¬ φ) ∧ ψ) ↔ ((φ ∧ ψ) ∨ (¬ φ ∧ ψ)))
4		iftrue 3669	. . . . 5 ⊢ (φ → if(φ, A, B) = A)
5		elimif.1	. . . . 5 ⊢ ( if(φ, A, B) = A → (ψ ↔ χ))
6	4, 5	syl 15	. . . 4 ⊢ (φ → (ψ ↔ χ))
7	6	pm5.32i 618	. . 3 ⊢ ((φ ∧ ψ) ↔ (φ ∧ χ))
8		iffalse 3670	. . . . 5 ⊢ (¬ φ → if(φ, A, B) = B)
9		elimif.2	. . . . 5 ⊢ ( if(φ, A, B) = B → (ψ ↔ θ))
10	8, 9	syl 15	. . . 4 ⊢ (¬ φ → (ψ ↔ θ))
11	10	pm5.32i 618	. . 3 ⊢ ((¬ φ ∧ ψ) ↔ (¬ φ ∧ θ))
12	7, 11	orbi12i 507	. 2 ⊢ (((φ ∧ ψ) ∨ (¬ φ ∧ ψ)) ↔ ((φ ∧ χ) ∨ (¬ φ ∧ θ)))
13	2, 3, 12	3bitri 262	1 ⊢ (ψ ↔ ((φ ∧ χ) ∨ (¬ φ ∧ θ)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 176   ∨ wo 357   ∧ wa 358   = wceq 1642   ifcif 3663

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-if 3664

This theorem is used by:  eqif  3696  elif  3697  ifel  3698