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Metamath Proof Explorer

Theorem frpoins2fg

Description: Well-Founded Induction schema, using implicit substitution. (Contributed by Scott Fenton, 24-Aug-2022)

Ref Expression
Hypotheses frpoins2fg.1 ⊢ ( 𝑦 ∈ 𝐴 → ( ∀ 𝑧 ∈ Pred ( 𝑅 , 𝐴 , 𝑦 ) 𝜓 → 𝜑 ) )
frpoins2fg.2 ⊢ Ⅎ 𝑦 𝜓
frpoins2fg.3 ⊢ ( 𝑦 = 𝑧 → ( 𝜑 ↔ 𝜓 ) )
Assertion frpoins2fg ( ( 𝑅 Fr 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴 ) → ∀ 𝑦 ∈ 𝐴 𝜑 )

Proof

Step Hyp Ref Expression
1 frpoins2fg.1 ⊢ ( 𝑦 ∈ 𝐴 → ( ∀ 𝑧 ∈ Pred ( 𝑅 , 𝐴 , 𝑦 ) 𝜓 → 𝜑 ) )
2 frpoins2fg.2 ⊢ Ⅎ 𝑦 𝜓
3 frpoins2fg.3 ⊢ ( 𝑦 = 𝑧 → ( 𝜑 ↔ 𝜓 ) )
4 sbsbc ⊢ ( [ 𝑧 / 𝑦 ] 𝜑 ↔ [ 𝑧 / 𝑦 ] 𝜑 )
5 2 3 sbiev ⊢ ( [ 𝑧 / 𝑦 ] 𝜑 ↔ 𝜓 )
6 4 5 bitr3i ⊢ ( [ 𝑧 / 𝑦 ] 𝜑 ↔ 𝜓 )
7 6 ralbii ⊢ ( ∀ 𝑧 ∈ Pred ( 𝑅 , 𝐴 , 𝑦 ) [ 𝑧 / 𝑦 ] 𝜑 ↔ ∀ 𝑧 ∈ Pred ( 𝑅 , 𝐴 , 𝑦 ) 𝜓 )
8 1 adantl ⊢ ( ( ( 𝑅 Fr 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴 ) ∧ 𝑦 ∈ 𝐴 ) → ( ∀ 𝑧 ∈ Pred ( 𝑅 , 𝐴 , 𝑦 ) 𝜓 → 𝜑 ) )
9 7 8 biimtrid ⊢ ( ( ( 𝑅 Fr 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴 ) ∧ 𝑦 ∈ 𝐴 ) → ( ∀ 𝑧 ∈ Pred ( 𝑅 , 𝐴 , 𝑦 ) [ 𝑧 / 𝑦 ] 𝜑 → 𝜑 ) )
10 9 frpoinsg ⊢ ( ( 𝑅 Fr 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴 ) → ∀ 𝑦 ∈ 𝐴 𝜑 )