Theorem ibllem 26140

Description: Conditioned equality theorem for the if statement. (Contributed by Mario Carneiro, 31-Jul-2014.)

Hypothesis

Ref	Expression
ibllem.1	⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝐵 = 𝐶)

Assertion

Ref	Expression
ibllem	⊢ (𝜑 → if((𝑥 ∈ 𝐴 ∧ 0 ≤ 𝐵), 𝐵, 0) = if((𝑥 ∈ 𝐴 ∧ 0 ≤ 𝐶), 𝐶, 0))

Proof of Theorem ibllem

Step	Hyp	Ref	Expression
1		ibllem.1	. . . . 5 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝐵 = 𝐶)
2	1	breq2d 5120	. . . 4 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (0 ≤ 𝐵 ↔ 0 ≤ 𝐶))
3	2	pm5.32da 590	. . 3 ⊢ (𝜑 → ((𝑥 ∈ 𝐴 ∧ 0 ≤ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 0 ≤ 𝐶)))
4	3	ifbid 4509	. 2 ⊢ (𝜑 → if((𝑥 ∈ 𝐴 ∧ 0 ≤ 𝐵), 𝐵, 0) = if((𝑥 ∈ 𝐴 ∧ 0 ≤ 𝐶), 𝐵, 0))
5	1	adantrr 730	. . 3 ⊢ ((𝜑 ∧ (𝑥 ∈ 𝐴 ∧ 0 ≤ 𝐶)) → 𝐵 = 𝐶)
6	5	ifeq1da 4517	. 2 ⊢ (𝜑 → if((𝑥 ∈ 𝐴 ∧ 0 ≤ 𝐶), 𝐵, 0) = if((𝑥 ∈ 𝐴 ∧ 0 ≤ 𝐶), 𝐶, 0))
7	4, 6	eqtrd 2798	1 ⊢ (𝜑 → if((𝑥 ∈ 𝐴 ∧ 0 ≤ 𝐵), 𝐵, 0) = if((𝑥 ∈ 𝐴 ∧ 0 ≤ 𝐶), 𝐶, 0))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1570   ∈ wcel 2147  ifcif 4485   class class class wbr 5108  0cc0 11239   ≤ cle 11383

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4841  df-br 5109

This theorem is used by:  isibl  26141  isibl2  26142  iblitg  26144  iblcnlem1  26163  iblcnlem  26164  itgcnlem  26165  iblrelem  26166  itgrevallem1  26170  itgeqa  26189