Theorem feq3 5521

Description: Equality theorem for functions. (Contributed by NM, 1-Aug-1994.)

Assertion

Ref	Expression
feq3	⊢ (𝐴 = 𝐵 → (𝐹:𝐶⟶𝐴 ↔ 𝐹:𝐶⟶𝐵))

Proof of Theorem feq3

Step	Hyp	Ref	Expression
1		sseq2 3272	. . 3 ⊢ (𝐴 = 𝐵 → (ran 𝐹 ⊆ 𝐴 ↔ ran 𝐹 ⊆ 𝐵))
2	1	anbi2d 468	. 2 ⊢ (𝐴 = 𝐵 → ((𝐹 Fn 𝐶 ∧ ran 𝐹 ⊆ 𝐴) ↔ (𝐹 Fn 𝐶 ∧ ran 𝐹 ⊆ 𝐵)))
3		df-f 5384	. 2 ⊢ (𝐹:𝐶⟶𝐴 ↔ (𝐹 Fn 𝐶 ∧ ran 𝐹 ⊆ 𝐴))
4		df-f 5384	. 2 ⊢ (𝐹:𝐶⟶𝐵 ↔ (𝐹 Fn 𝐶 ∧ ran 𝐹 ⊆ 𝐵))
5	2, 3, 4	3bitr4g 223	1 ⊢ (𝐴 = 𝐵 → (𝐹:𝐶⟶𝐴 ↔ 𝐹:𝐶⟶𝐵))

This proof depends on syntax axioms:   → wi 4   ∧ wa 104   ↔ wb 105   = wceq 1402   ⊆ wss 3220  ran crn 4777   Fn wfn 5375  ⟶wf 5376

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-11 1559  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-in 3226  df-ss 3233  df-f 5384

This theorem is used by:  feq23  5522  feq3d  5525  feq123d  5527  fun2  5565  fconstg  5592  f1eq3  5598  fsng  5884  fsn2  5885  fsnunf  5918  mapvalg  6936  mapsnd  6974  mapsn  6976  lmff  15475  txcn  15501  plyrecj  15989  umgrislfupgrdom  16577  uspgriedgedg  16625  usgrislfuspgrdom  16636  subupgr  16719  wlkv0  16815