sbied

Description: Conversion of implicit substitution to explicit substitution (deduction version of sbie ) Usage of this theorem is discouraged because it depends on ax-13 . See sbiedw , sbiedvw for variants using disjoint variables, but requiring fewer axioms. (Contributed by NM, 30-Jun-1994) (Revised by Mario Carneiro, 4-Oct-2016) (Proof shortened by Wolf Lammen, 24-Jun-2018) (New usage is discouraged.)

	Ref	Expression
Hypotheses	sbied.1	⊢ Ⅎ 𝑥 𝜑
	sbied.2	⊢ ( 𝜑 → Ⅎ 𝑥 𝜒 )
	sbied.3	⊢ ( 𝜑 → ( 𝑥 = 𝑦 → ( 𝜓 ↔ 𝜒 ) ) )
Assertion	sbied	⊢ ( 𝜑 → ( [ 𝑦 / 𝑥 ] 𝜓 ↔ 𝜒 ) )

Proof

Step	Hyp	Ref	Expression
1		sbied.1	⊢ Ⅎ 𝑥 𝜑
2		sbied.2	⊢ ( 𝜑 → Ⅎ 𝑥 𝜒 )
3		sbied.3	⊢ ( 𝜑 → ( 𝑥 = 𝑦 → ( 𝜓 ↔ 𝜒 ) ) )
4	1	sbrim	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 → 𝜓 ) ↔ ( 𝜑 → [ 𝑦 / 𝑥 ] 𝜓 ) )
5	1 2	nfim1	⊢ Ⅎ 𝑥 ( 𝜑 → 𝜒 )
6	3	com12	⊢ ( 𝑥 = 𝑦 → ( 𝜑 → ( 𝜓 ↔ 𝜒 ) ) )
7	6	pm5.74d	⊢ ( 𝑥 = 𝑦 → ( ( 𝜑 → 𝜓 ) ↔ ( 𝜑 → 𝜒 ) ) )
8	5 7	sbie	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 → 𝜓 ) ↔ ( 𝜑 → 𝜒 ) )
9	4 8	bitr3i	⊢ ( ( 𝜑 → [ 𝑦 / 𝑥 ] 𝜓 ) ↔ ( 𝜑 → 𝜒 ) )
10	9	pm5.74ri	⊢ ( 𝜑 → ( [ 𝑦 / 𝑥 ] 𝜓 ↔ 𝜒 ) )

Metamath Proof Explorer

Theorem sbied

Proof