Theorem funbreq 35770

Description: An equality condition for functions. (Contributed by Scott Fenton, 18-Feb-2013.)

Hypotheses

Ref	Expression
funbreq.1	⊢ 𝐴 ∈ V
funbreq.2	⊢ 𝐵 ∈ V
funbreq.3	⊢ 𝐶 ∈ V

Assertion

Ref	Expression
funbreq	⊢ ((Fun 𝐹 ∧ 𝐴𝐹𝐵) → (𝐴𝐹𝐶 ↔ 𝐵 = 𝐶))

Proof of Theorem funbreq

Step	Hyp	Ref	Expression
1		funbreq.1	. . . 4 ⊢ 𝐴 ∈ V
2		funbreq.2	. . . 4 ⊢ 𝐵 ∈ V
3		funbreq.3	. . . 4 ⊢ 𝐶 ∈ V
4	1, 2, 3	fununiq 35769	. . 3 ⊢ (Fun 𝐹 → ((𝐴𝐹𝐵 ∧ 𝐴𝐹𝐶) → 𝐵 = 𝐶))
5	4	expdimp 452	. 2 ⊢ ((Fun 𝐹 ∧ 𝐴𝐹𝐵) → (𝐴𝐹𝐶 → 𝐵 = 𝐶))
6		breq2 5147	. . . 4 ⊢ (𝐵 = 𝐶 → (𝐴𝐹𝐵 ↔ 𝐴𝐹𝐶))
7	6	biimpcd 249	. . 3 ⊢ (𝐴𝐹𝐵 → (𝐵 = 𝐶 → 𝐴𝐹𝐶))
8	7	adantl 481	. 2 ⊢ ((Fun 𝐹 ∧ 𝐴𝐹𝐵) → (𝐵 = 𝐶 → 𝐴𝐹𝐶))
9	5, 8	impbid 212	1 ⊢ ((Fun 𝐹 ∧ 𝐴𝐹𝐵) → (𝐴𝐹𝐶 ↔ 𝐵 = 𝐶))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1540   ∈ wcel 2108  Vcvv 3480   class class class wbr 5143  Fun wfun 6555

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2708  ax-sep 5296  ax-nul 5306  ax-pr 5432

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-ral 3062  df-rex 3071  df-rab 3437  df-v 3482  df-dif 3954  df-un 3956  df-ss 3968  df-nul 4334  df-if 4526  df-sn 4627  df-pr 4629  df-op 4633  df-br 5144  df-opab 5206  df-id 5578  df-xp 5691  df-rel 5692  df-cnv 5693  df-co 5694  df-fun 6563

This theorem is referenced by: (None)