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Metamath Proof Explorer

Theorem sbhypf

Description: Introduce an explicit substitution into an implicit substitution hypothesis. See also csbhypf . (Contributed by Raph Levien, 10-Apr-2004) (Proof shortened by Wolf Lammen, 25-Jan-2025)

	Ref	Expression
Hypotheses	sbhypf.1	⊢ Ⅎ 𝑥 𝜓
	sbhypf.2	⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
Assertion	sbhypf	⊢ ( 𝑦 = 𝐴 → ( [ 𝑦 / 𝑥 ] 𝜑 ↔ 𝜓 ) )

Proof

Step	Hyp	Ref	Expression
1		sbhypf.1	⊢ Ⅎ 𝑥 𝜓
2		sbhypf.2	⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
3	2	sbimi	⊢ ( [ 𝑦 / 𝑥 ] 𝑥 = 𝐴 → [ 𝑦 / 𝑥 ] ( 𝜑 ↔ 𝜓 ) )
4		eqsb1	⊢ ( [ 𝑦 / 𝑥 ] 𝑥 = 𝐴 ↔ 𝑦 = 𝐴 )
5	1	sbf	⊢ ( [ 𝑦 / 𝑥 ] 𝜓 ↔ 𝜓 )
6	5	sblbis	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ↔ 𝜓 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ 𝜓 ) )
7	3 4 6	3imtr3i	⊢ ( 𝑦 = 𝐴 → ( [ 𝑦 / 𝑥 ] 𝜑 ↔ 𝜓 ) )