Metamath Proof Explorer


Theorem poeq2

Description: Equality theorem for partial ordering predicate. (Contributed by NM, 27-Mar-1997)

Ref Expression
Assertion poeq2 ( 𝐴 = 𝐵 → ( 𝑅 Po 𝐴 ↔ 𝑅 Po 𝐵 ) )

Proof

Step Hyp Ref Expression
1 eqimss2 ⊢ ( 𝐴 = 𝐵 → 𝐵 ⊆ 𝐴 )
2 poss ⊢ ( 𝐵 ⊆ 𝐴 → ( 𝑅 Po 𝐴 → 𝑅 Po 𝐵 ) )
3 1 2 syl ⊢ ( 𝐴 = 𝐵 → ( 𝑅 Po 𝐴 → 𝑅 Po 𝐵 ) )
4 eqimss ⊢ ( 𝐴 = 𝐵 → 𝐴 ⊆ 𝐵 )
5 poss ⊢ ( 𝐴 ⊆ 𝐵 → ( 𝑅 Po 𝐵 → 𝑅 Po 𝐴 ) )
6 4 5 syl ⊢ ( 𝐴 = 𝐵 → ( 𝑅 Po 𝐵 → 𝑅 Po 𝐴 ) )
7 3 6 impbid ⊢ ( 𝐴 = 𝐵 → ( 𝑅 Po 𝐴 ↔ 𝑅 Po 𝐵 ) )