Theorem funbreq 35733

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 35732	. . 3 ⊢ (Fun 𝐹 → ((𝐴𝐹𝐵 ∧ 𝐴𝐹𝐶) → 𝐵 = 𝐶))
5	4	expdimp 452	. 2 ⊢ ((Fun 𝐹 ∧ 𝐴𝐹𝐵) → (𝐴𝐹𝐶 → 𝐵 = 𝐶))
6		breq2 5170	. . . 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 1537   ∈ wcel 2108  Vcvv 3488   class class class wbr 5166  Fun wfun 6567

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711  ax-sep 5317  ax-nul 5324  ax-pr 5447

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-3an 1089  df-tru 1540  df-fal 1550  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-ral 3068  df-rex 3077  df-rab 3444  df-v 3490  df-dif 3979  df-un 3981  df-ss 3993  df-nul 4353  df-if 4549  df-sn 4649  df-pr 4651  df-op 4655  df-br 5167  df-opab 5229  df-id 5593  df-xp 5706  df-rel 5707  df-cnv 5708  df-co 5709  df-fun 6575

This theorem is referenced by: (None)