Theorem poeq12d 5563

Description: Equality deduction for partial orderings. (Contributed by Matthew House, 10-Sep-2025.)

Hypotheses

Ref	Expression
poeq12d.1	⊢ (𝜑 → 𝑅 = 𝑆)
poeq12d.2	⊢ (𝜑 → 𝐴 = 𝐵)

Assertion

Ref	Expression
poeq12d	⊢ (𝜑 → (𝑅 Po 𝐴 ↔ 𝑆 Po 𝐵))

Proof of Theorem poeq12d

Step	Hyp	Ref	Expression
1		poeq12d.1	. 2 ⊢ (𝜑 → 𝑅 = 𝑆)
2		poeq12d.2	. 2 ⊢ (𝜑 → 𝐴 = 𝐵)
3		poeq1 5561	. . 3 ⊢ (𝑅 = 𝑆 → (𝑅 Po 𝐴 ↔ 𝑆 Po 𝐴))
4		poeq2 5562	. . 3 ⊢ (𝐴 = 𝐵 → (𝑆 Po 𝐴 ↔ 𝑆 Po 𝐵))
5	3, 4	sylan9bb 519	. 2 ⊢ ((𝑅 = 𝑆 ∧ 𝐴 = 𝐵) → (𝑅 Po 𝐴 ↔ 𝑆 Po 𝐵))
6	1, 2, 5	syl2anc 596	1 ⊢ (𝜑 → (𝑅 Po 𝐴 ↔ 𝑆 Po 𝐵))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1569   Po wpo 5556

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-cleq 2752  df-clel 2835  df-ral 3077  df-ss 3896  df-br 5103  df-po 5558

This theorem is used by:  weiunpo  37295