Theorem sseq2d 3027

Description: An equality deduction for the subclass relationship. (Contributed by NM, 14-Aug-1994.)

Hypothesis

Ref	Expression
sseq1d.1	⊢ (𝜑 → 𝐴 = 𝐵)

Assertion

Ref	Expression
sseq2d	⊢ (𝜑 → (𝐶 ⊆ 𝐴 ↔ 𝐶 ⊆ 𝐵))

Proof of Theorem sseq2d

Step	Hyp	Ref	Expression
1		sseq1d.1	. 2 ⊢ (𝜑 → 𝐴 = 𝐵)
2		sseq2 3021	. 2 ⊢ (𝐴 = 𝐵 → (𝐶 ⊆ 𝐴 ↔ 𝐶 ⊆ 𝐵))
3	1, 2	syl 14	1 ⊢ (𝜑 → (𝐶 ⊆ 𝐴 ↔ 𝐶 ⊆ 𝐵))

Syntax hints:   → wi 4   ↔ wb 103   = wceq 1284   ⊆ wss 2973

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-11 1437  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-in 2979  df-ss 2986

This theorem is referenced by:  sseq12d  3028  sseqtrd  3035  onsucsssucexmid  4270  sbcrel  4444  funimass2  4997  fnco  5027  fnssresb  5031  fnimaeq0  5040  foimacnv  5164  fvelimab  5250  ssimaexg  5256  fvmptss2  5268  rdgss  5993  frecsuclemdm  6011