Theorem funbreq 36262

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 36261	. . 3 ⊢ (Fun 𝐹 → ((𝐴𝐹𝐵 ∧ 𝐴𝐹𝐶) → 𝐵 = 𝐶))
5	4	expdimp 457	. 2 ⊢ ((Fun 𝐹 ∧ 𝐴𝐹𝐵) → (𝐴𝐹𝐶 → 𝐵 = 𝐶))
6		breq2 5113	. . . 4 ⊢ (𝐵 = 𝐶 → (𝐴𝐹𝐵 ↔ 𝐴𝐹𝐶))
7	6	biimpcd 252	. . 3 ⊢ (𝐴𝐹𝐵 → (𝐵 = 𝐶 → 𝐴𝐹𝐶))
8	7	adantl 486	. 2 ⊢ ((Fun 𝐹 ∧ 𝐴𝐹𝐵) → (𝐵 = 𝐶 → 𝐴𝐹𝐶))
9	5, 8	impbid 215	1 ⊢ ((Fun 𝐹 ∧ 𝐴𝐹𝐵) → (𝐴𝐹𝐶 ↔ 𝐵 = 𝐶))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 400   = wceq 1570   ∈ wcel 2143  Vcvv 3455   class class class wbr 5109  Fun wfun 6530

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735  ax-sep 5257  ax-pr 5404

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-ral 3080  df-rex 3090  df-rab 3417  df-v 3457  df-dif 3908  df-un 3910  df-in 3912  df-ss 3922  df-nul 4287  df-if 4488  df-sn 4590  df-pr 4592  df-op 4596  df-br 5110  df-opab 5174  df-id 5556  df-xp 5667  df-rel 5668  df-cnv 5669  df-co 5670  df-fun 6538

This theorem is referenced by: (None)