Theorem ordeq 4109

Description: Equality theorem for the ordinal predicate. (Contributed by NM, 17-Sep-1993.)

Assertion

Ref	Expression
ordeq	⊢ (𝐴 = 𝐵 → (Ord 𝐴 ↔ Ord 𝐵))

Proof of Theorem ordeq

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		treq 3860	. . 3 ⊢ (𝐴 = 𝐵 → (Tr 𝐴 ↔ Tr 𝐵))
2		raleq 2505	. . 3 ⊢ (𝐴 = 𝐵 → (∀𝑥 ∈ 𝐴 Tr 𝑥 ↔ ∀𝑥 ∈ 𝐵 Tr 𝑥))
3	1, 2	anbi12d 442	. 2 ⊢ (𝐴 = 𝐵 → ((Tr 𝐴 ∧ ∀𝑥 ∈ 𝐴 Tr 𝑥) ↔ (Tr 𝐵 ∧ ∀𝑥 ∈ 𝐵 Tr 𝑥)))
4		dford3 4104	. 2 ⊢ (Ord 𝐴 ↔ (Tr 𝐴 ∧ ∀𝑥 ∈ 𝐴 Tr 𝑥))
5		dford3 4104	. 2 ⊢ (Ord 𝐵 ↔ (Tr 𝐵 ∧ ∀𝑥 ∈ 𝐵 Tr 𝑥))
6	3, 4, 5	3bitr4g 212	1 ⊢ (𝐴 = 𝐵 → (Ord 𝐴 ↔ Ord 𝐵))

Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98   = wceq 1243  ∀wral 2306  Tr wtr 3854  Ord word 4099

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-ral 2311  df-rex 2312  df-in 2924  df-ss 2931  df-uni 3581  df-tr 3855  df-iord 4103

This theorem is referenced by:  elong  4110  limeq  4114  ordelord  4118  ordtriexmidlem  4245  2ordpr  4249  issmo  5903  issmo2  5904  smoeq  5905  smores  5907  smores2  5909  smodm2  5910  smoiso  5917  tfrlem8  5934