Metamath Proof Explorer


Theorem frege80

Description: Add additional condition to both clauses of frege79 . Proposition 80 of Frege1879 p. 63. (Contributed by RP, 1-Jul-2020) (Revised by RP, 5-Jul-2020) (Proof modification is discouraged.)

Ref Expression
Hypotheses frege80.x ⊢ 𝑋 ∈ 𝑈
frege80.y ⊢ 𝑌 ∈ 𝑉
frege80.r ⊢ 𝑅 ∈ 𝑊
frege80.a ⊢ 𝐴 ∈ 𝐵
Assertion frege80 ( ( 𝑋 ∈ 𝐴 → ( 𝑅 hereditary 𝐴 → ∀ 𝑎 ( 𝑋 𝑅 𝑎 → 𝑎 ∈ 𝐴 ) ) ) → ( 𝑋 ∈ 𝐴 → ( 𝑅 hereditary 𝐴 → ( 𝑋 ( t+ ‘ 𝑅 ) 𝑌 → 𝑌 ∈ 𝐴 ) ) ) )

Proof

Step Hyp Ref Expression
1 frege80.x ⊢ 𝑋 ∈ 𝑈
2 frege80.y ⊢ 𝑌 ∈ 𝑉
3 frege80.r ⊢ 𝑅 ∈ 𝑊
4 frege80.a ⊢ 𝐴 ∈ 𝐵
5 1 2 3 4 frege79 ⊢ ( ( 𝑅 hereditary 𝐴 → ∀ 𝑎 ( 𝑋 𝑅 𝑎 → 𝑎 ∈ 𝐴 ) ) → ( 𝑅 hereditary 𝐴 → ( 𝑋 ( t+ ‘ 𝑅 ) 𝑌 → 𝑌 ∈ 𝐴 ) ) )
6 frege5 ⊢ ( ( ( 𝑅 hereditary 𝐴 → ∀ 𝑎 ( 𝑋 𝑅 𝑎 → 𝑎 ∈ 𝐴 ) ) → ( 𝑅 hereditary 𝐴 → ( 𝑋 ( t+ ‘ 𝑅 ) 𝑌 → 𝑌 ∈ 𝐴 ) ) ) → ( ( 𝑋 ∈ 𝐴 → ( 𝑅 hereditary 𝐴 → ∀ 𝑎 ( 𝑋 𝑅 𝑎 → 𝑎 ∈ 𝐴 ) ) ) → ( 𝑋 ∈ 𝐴 → ( 𝑅 hereditary 𝐴 → ( 𝑋 ( t+ ‘ 𝑅 ) 𝑌 → 𝑌 ∈ 𝐴 ) ) ) ) )
7 5 6 ax-mp ⊢ ( ( 𝑋 ∈ 𝐴 → ( 𝑅 hereditary 𝐴 → ∀ 𝑎 ( 𝑋 𝑅 𝑎 → 𝑎 ∈ 𝐴 ) ) ) → ( 𝑋 ∈ 𝐴 → ( 𝑅 hereditary 𝐴 → ( 𝑋 ( t+ ‘ 𝑅 ) 𝑌 → 𝑌 ∈ 𝐴 ) ) ) )