Metamath Proof Explorer


Theorem sbiedw

Description: Conversion of implicit substitution to explicit substitution (deduction version of sbiev ). Version of sbied with a disjoint variable condition, requiring fewer axioms. (Contributed by NM, 30-Jun-1994) Avoid ax-13 . (Revised by GG, 10-Jan-2024)

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

Proof

Step Hyp Ref Expression
1 sbiedw.1 ⊢ Ⅎ 𝑥 𝜑
2 sbiedw.2 ⊢ ( 𝜑 → Ⅎ 𝑥 𝜒 )
3 sbiedw.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 sbiev ⊢ ( [ 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 ) )