Metamath Proof Explorer


Theorem poeq12d

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

Ref Expression
Hypotheses poeq12d.1 φ R = S
poeq12d.2 φ A = B
Assertion poeq12d φ R Po A S Po B

Proof

Step Hyp Ref Expression
1 poeq12d.1 φ R = S
2 poeq12d.2 φ A = B
3 poeq1 R = S R Po A S Po A
4 poeq2 A = B S Po A S Po B
5 3 4 sylan9bb R = S A = B R Po A S Po B
6 1 2 5 syl2anc φ R Po A S Po B