Theorem funbreq 36658

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 6560	. . 3 ⊢ (Fun 𝐹 → ((𝐴𝐹𝐵 ∧ 𝐴𝐹𝐶) → 𝐵 = 𝐶))
5	4	expdimp 458	. 2 ⊢ ((Fun 𝐹 ∧ 𝐴𝐹𝐵) → (𝐴𝐹𝐶 → 𝐵 = 𝐶))
6		breq2 5106	. . . 4 ⊢ (𝐵 = 𝐶 → (𝐴𝐹𝐵 ↔ 𝐴𝐹𝐶))
7	6	biimpcd 252	. . 3 ⊢ (𝐴𝐹𝐵 → (𝐵 = 𝐶 → 𝐴𝐹𝐶))
8	7	adantl 487	. 2 ⊢ ((Fun 𝐹 ∧ 𝐴𝐹𝐵) → (𝐵 = 𝐶 → 𝐴𝐹𝐶))
9	5, 8	impbid 215	1 ⊢ ((Fun 𝐹 ∧ 𝐴𝐹𝐵) → (𝐴𝐹𝐶 ↔ 𝐵 = 𝐶))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1569   ∈ wcel 2144  Vcvv 3450   class class class wbr 5102  Fun wfun 6537

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732  ax-sep 5248  ax-pr 5391

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-ss 3897  df-dif 4046  df-in 4090  df-un 4150  df-nul 4278  df-if 4482  df-sn 4585  df-pr 4587  df-op 4835  df-br 5103  df-opab 5167  df-id 5546  df-xp 5657  df-rel 5658  df-cnv 5659  df-co 5660  df-fun 6545

This theorem is used by: (None)