Theorem funbreq 31668

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 31667	. . 3 ⊢ (Fun 𝐹 → ((𝐴𝐹𝐵 ∧ 𝐴𝐹𝐶) → 𝐵 = 𝐶))
5	4	expdimp 453	. 2 ⊢ ((Fun 𝐹 ∧ 𝐴𝐹𝐵) → (𝐴𝐹𝐶 → 𝐵 = 𝐶))
6		breq2 4657	. . . 4 ⊢ (𝐵 = 𝐶 → (𝐴𝐹𝐵 ↔ 𝐴𝐹𝐶))
7	6	biimpcd 239	. . 3 ⊢ (𝐴𝐹𝐵 → (𝐵 = 𝐶 → 𝐴𝐹𝐶))
8	7	adantl 482	. 2 ⊢ ((Fun 𝐹 ∧ 𝐴𝐹𝐵) → (𝐵 = 𝐶 → 𝐴𝐹𝐶))
9	5, 8	impbid 202	1 ⊢ ((Fun 𝐹 ∧ 𝐴𝐹𝐵) → (𝐴𝐹𝐶 ↔ 𝐵 = 𝐶))

Syntax hints:   → wi 4   ↔ wb 196   ∧ wa 384   = wceq 1483   ∈ wcel 1990  Vcvv 3200   class class class wbr 4653  Fun wfun 5882

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602  ax-sep 4781  ax-nul 4789  ax-pr 4906

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1039  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-eu 2474  df-mo 2475  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-ral 2917  df-rab 2921  df-v 3202  df-dif 3577  df-un 3579  df-in 3581  df-ss 3588  df-nul 3916  df-if 4087  df-sn 4178  df-pr 4180  df-op 4184  df-br 4654  df-opab 4713  df-id 5024  df-cnv 5122  df-co 5123  df-fun 5890

This theorem is referenced by: (None)