Metamath Proof Explorer


Theorem ss2abdv

Description: Deduction of abstraction subclass from implication. (Contributed by NM, 29-Jul-2011) Reduce dependencies on axioms. (Revised by Steven Nguyen, 28-Jun-2024)

Ref Expression
Hypothesis ss2abdv.1 ⊢ φ → ψ → χ
Assertion ss2abdv ⊢ φ → x | ψ ⊆ x | χ

Proof

Step Hyp Ref Expression
1 ss2abdv.1 ⊢ φ → ψ → χ
2 1 sbimdv ⊢ φ → y x ψ → y x χ
3 df-clab ⊢ y ∈ x | ψ ↔ y x ψ
4 df-clab ⊢ y ∈ x | χ ↔ y x χ
5 2 3 4 3imtr4g ⊢ φ → y ∈ x | ψ → y ∈ x | χ
6 5 ssrdv ⊢ φ → x | ψ ⊆ x | χ