Metamath Proof Explorer


Theorem bnj1147

Description: Property of _trCl . (Contributed by Jonathan Ben-Naim, 3-Jun-2011) (New usage is discouraged.)

Ref Expression
Assertion bnj1147 trCl ( 𝑋 , 𝐴 , 𝑅 ) ⊆ 𝐴

Proof

Step Hyp Ref Expression
1 biid ( ( 𝑓 ‘ ∅ ) = pred ( 𝑋 , 𝐴 , 𝑅 ) ↔ ( 𝑓 ‘ ∅ ) = pred ( 𝑋 , 𝐴 , 𝑅 ) )
2 biid ( ∀ 𝑖 ∈ ω ( suc 𝑖𝑛 → ( 𝑓 ‘ suc 𝑖 ) = 𝑦 ∈ ( 𝑓𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) ↔ ∀ 𝑖 ∈ ω ( suc 𝑖𝑛 → ( 𝑓 ‘ suc 𝑖 ) = 𝑦 ∈ ( 𝑓𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) )
3 eqid ( ω ∖ { ∅ } ) = ( ω ∖ { ∅ } )
4 eqid { 𝑓 ∣ ∃ 𝑛 ∈ ( ω ∖ { ∅ } ) ( 𝑓 Fn 𝑛 ∧ ( 𝑓 ‘ ∅ ) = pred ( 𝑋 , 𝐴 , 𝑅 ) ∧ ∀ 𝑖 ∈ ω ( suc 𝑖𝑛 → ( 𝑓 ‘ suc 𝑖 ) = 𝑦 ∈ ( 𝑓𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) ) } = { 𝑓 ∣ ∃ 𝑛 ∈ ( ω ∖ { ∅ } ) ( 𝑓 Fn 𝑛 ∧ ( 𝑓 ‘ ∅ ) = pred ( 𝑋 , 𝐴 , 𝑅 ) ∧ ∀ 𝑖 ∈ ω ( suc 𝑖𝑛 → ( 𝑓 ‘ suc 𝑖 ) = 𝑦 ∈ ( 𝑓𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) ) }
5 biid ( ( 𝑛 ∈ ( ω ∖ { ∅ } ) ∧ 𝑓 Fn 𝑛 ∧ ( 𝑓 ‘ ∅ ) = pred ( 𝑋 , 𝐴 , 𝑅 ) ∧ ∀ 𝑖 ∈ ω ( suc 𝑖𝑛 → ( 𝑓 ‘ suc 𝑖 ) = 𝑦 ∈ ( 𝑓𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) ) ↔ ( 𝑛 ∈ ( ω ∖ { ∅ } ) ∧ 𝑓 Fn 𝑛 ∧ ( 𝑓 ‘ ∅ ) = pred ( 𝑋 , 𝐴 , 𝑅 ) ∧ ∀ 𝑖 ∈ ω ( suc 𝑖𝑛 → ( 𝑓 ‘ suc 𝑖 ) = 𝑦 ∈ ( 𝑓𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) ) )
6 biid ( ( ( 𝑖 ≠ ∅ ∧ 𝑖𝑛 ∧ ( 𝑛 ∈ ( ω ∖ { ∅ } ) ∧ 𝑓 Fn 𝑛 ∧ ( 𝑓 ‘ ∅ ) = pred ( 𝑋 , 𝐴 , 𝑅 ) ∧ ∀ 𝑖 ∈ ω ( suc 𝑖𝑛 → ( 𝑓 ‘ suc 𝑖 ) = 𝑦 ∈ ( 𝑓𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) ) ) ∧ ( 𝑗𝑛𝑖 = suc 𝑗 ) ) ↔ ( ( 𝑖 ≠ ∅ ∧ 𝑖𝑛 ∧ ( 𝑛 ∈ ( ω ∖ { ∅ } ) ∧ 𝑓 Fn 𝑛 ∧ ( 𝑓 ‘ ∅ ) = pred ( 𝑋 , 𝐴 , 𝑅 ) ∧ ∀ 𝑖 ∈ ω ( suc 𝑖𝑛 → ( 𝑓 ‘ suc 𝑖 ) = 𝑦 ∈ ( 𝑓𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) ) ) ∧ ( 𝑗𝑛𝑖 = suc 𝑗 ) ) )
7 1 2 3 4 5 6 bnj1145 trCl ( 𝑋 , 𝐴 , 𝑅 ) ⊆ 𝐴