Metamath Proof Explorer


Theorem sbcom

Description: A commutativity law for substitution. Usage of this theorem is discouraged because it depends on ax-13 . Check out sbcom3vv for a version requiring fewer axioms. (Contributed by NM, 27-May-1997) (Proof shortened by Wolf Lammen, 20-Sep-2018) (New usage is discouraged.)

Ref Expression
Assertion sbcom ( [ 𝑦 / 𝑧 ] [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝑦 / 𝑥 ] [ 𝑦 / 𝑧 ] 𝜑 )

Proof

Step Hyp Ref Expression
1 sbco3 ( [ 𝑦 / 𝑧 ] [ 𝑧 / 𝑥 ] 𝜑 ↔ [ 𝑦 / 𝑥 ] [ 𝑥 / 𝑧 ] 𝜑 )
2 sbcom3 ( [ 𝑦 / 𝑧 ] [ 𝑧 / 𝑥 ] 𝜑 ↔ [ 𝑦 / 𝑧 ] [ 𝑦 / 𝑥 ] 𝜑 )
3 sbcom3 ( [ 𝑦 / 𝑥 ] [ 𝑥 / 𝑧 ] 𝜑 ↔ [ 𝑦 / 𝑥 ] [ 𝑦 / 𝑧 ] 𝜑 )
4 1 2 3 3bitr3i ( [ 𝑦 / 𝑧 ] [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝑦 / 𝑥 ] [ 𝑦 / 𝑧 ] 𝜑 )