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Metamath Proof Explorer

Theorem sbco

Description: A composition law for substitution. Usage of this theorem is discouraged because it depends on ax-13 . See sbcov for a version with a disjoint variable condition requiring fewer axioms. (Contributed by NM, 14-May-1993) (Proof shortened by Wolf Lammen, 21-Sep-2018) (New usage is discouraged.)

		Ref	Expression
	Assertion	sbco	⊢ ( [ 𝑦 / 𝑥 ] [ 𝑥 / 𝑦 ] 𝜑 ↔ [ 𝑦 / 𝑥 ] 𝜑 )

Proof

Step	Hyp	Ref	Expression
1		sbcom3	⊢ ( [ 𝑦 / 𝑥 ] [ 𝑥 / 𝑦 ] 𝜑 ↔ [ 𝑦 / 𝑥 ] [ 𝑦 / 𝑦 ] 𝜑 )
2		sbid	⊢ ( [ 𝑦 / 𝑦 ] 𝜑 ↔ 𝜑 )
3	2	sbbii	⊢ ( [ 𝑦 / 𝑥 ] [ 𝑦 / 𝑦 ] 𝜑 ↔ [ 𝑦 / 𝑥 ] 𝜑 )
4	1 3	bitri	⊢ ( [ 𝑦 / 𝑥 ] [ 𝑥 / 𝑦 ] 𝜑 ↔ [ 𝑦 / 𝑥 ] 𝜑 )