Theorem suceqd 6399

Description: Deduction associated with suceq 6400. (Contributed by Rohan Ridenour, 8-Aug-2023.)

Hypothesis

Ref	Expression
suceqd.1	⊢ (𝜑 → 𝐴 = 𝐵)

Assertion

Ref	Expression
suceqd	⊢ (𝜑 → suc 𝐴 = suc 𝐵)

Proof of Theorem suceqd

Step	Hyp	Ref	Expression
1		suceqd.1	. . 3 ⊢ (𝜑 → 𝐴 = 𝐵)
2	1	sneqd 4601	. . 3 ⊢ (𝜑 → {𝐴} = {𝐵})
3	1, 2	uneq12d 4132	. 2 ⊢ (𝜑 → (𝐴 ∪ {𝐴}) = (𝐵 ∪ {𝐵}))
4		df-suc 6338	. 2 ⊢ suc 𝐴 = (𝐴 ∪ {𝐴})
5		df-suc 6338	. 2 ⊢ suc 𝐵 = (𝐵 ∪ {𝐵})
6	3, 4, 5	3eqtr4g 2789	1 ⊢ (𝜑 → suc 𝐴 = suc 𝐵)

Syntax hints:   → wi 4   = wceq 1540   ∪ cun 3912  {csn 4589  suc csuc 6334

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2701

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2721  df-clel 2803  df-v 3449  df-un 3919  df-sn 4590  df-suc 6338

This theorem is referenced by:  suceq  6400  nosupbnd2  27628  scottrankd  44237