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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 )