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Metamath Proof Explorer

Theorem ttrcleq

Description: Equality theorem for transitive closure. (Contributed by Scott Fenton, 17-Oct-2024)

Ref Expression
Assertion ttrcleq ( 𝑅 = 𝑆 → t++ 𝑅 = t++ 𝑆 )

Proof

Step Hyp Ref Expression
1 breq ⊢ ( 𝑅 = 𝑆 → ( ( 𝑓 ‘ 𝑚 ) 𝑅 ( 𝑓 ‘ suc 𝑚 ) ↔ ( 𝑓 ‘ 𝑚 ) 𝑆 ( 𝑓 ‘ suc 𝑚 ) ) )
2 1 ralbidv ⊢ ( 𝑅 = 𝑆 → ( ∀ 𝑚 ∈ 𝑛 ( 𝑓 ‘ 𝑚 ) 𝑅 ( 𝑓 ‘ suc 𝑚 ) ↔ ∀ 𝑚 ∈ 𝑛 ( 𝑓 ‘ 𝑚 ) 𝑆 ( 𝑓 ‘ suc 𝑚 ) ) )
3 2 3anbi3d ⊢ ( 𝑅 = 𝑆 → ( ( 𝑓 Fn suc 𝑛 ∧ ( ( 𝑓 ‘ ∅ ) = 𝑥 ∧ ( 𝑓 ‘ 𝑛 ) = 𝑦 ) ∧ ∀ 𝑚 ∈ 𝑛 ( 𝑓 ‘ 𝑚 ) 𝑅 ( 𝑓 ‘ suc 𝑚 ) ) ↔ ( 𝑓 Fn suc 𝑛 ∧ ( ( 𝑓 ‘ ∅ ) = 𝑥 ∧ ( 𝑓 ‘ 𝑛 ) = 𝑦 ) ∧ ∀ 𝑚 ∈ 𝑛 ( 𝑓 ‘ 𝑚 ) 𝑆 ( 𝑓 ‘ suc 𝑚 ) ) ) )
4 3 exbidv ⊢ ( 𝑅 = 𝑆 → ( ∃ 𝑓 ( 𝑓 Fn suc 𝑛 ∧ ( ( 𝑓 ‘ ∅ ) = 𝑥 ∧ ( 𝑓 ‘ 𝑛 ) = 𝑦 ) ∧ ∀ 𝑚 ∈ 𝑛 ( 𝑓 ‘ 𝑚 ) 𝑅 ( 𝑓 ‘ suc 𝑚 ) ) ↔ ∃ 𝑓 ( 𝑓 Fn suc 𝑛 ∧ ( ( 𝑓 ‘ ∅ ) = 𝑥 ∧ ( 𝑓 ‘ 𝑛 ) = 𝑦 ) ∧ ∀ 𝑚 ∈ 𝑛 ( 𝑓 ‘ 𝑚 ) 𝑆 ( 𝑓 ‘ suc 𝑚 ) ) ) )
5 4 rexbidv ⊢ ( 𝑅 = 𝑆 → ( ∃ 𝑛 ∈ ( ω ∖ 1o ) ∃ 𝑓 ( 𝑓 Fn suc 𝑛 ∧ ( ( 𝑓 ‘ ∅ ) = 𝑥 ∧ ( 𝑓 ‘ 𝑛 ) = 𝑦 ) ∧ ∀ 𝑚 ∈ 𝑛 ( 𝑓 ‘ 𝑚 ) 𝑅 ( 𝑓 ‘ suc 𝑚 ) ) ↔ ∃ 𝑛 ∈ ( ω ∖ 1o ) ∃ 𝑓 ( 𝑓 Fn suc 𝑛 ∧ ( ( 𝑓 ‘ ∅ ) = 𝑥 ∧ ( 𝑓 ‘ 𝑛 ) = 𝑦 ) ∧ ∀ 𝑚 ∈ 𝑛 ( 𝑓 ‘ 𝑚 ) 𝑆 ( 𝑓 ‘ suc 𝑚 ) ) ) )
6 5 opabbidv ⊢ ( 𝑅 = 𝑆 → { ⟨ 𝑥 , 𝑦 ⟩ ∣ ∃ 𝑛 ∈ ( ω ∖ 1o ) ∃ 𝑓 ( 𝑓 Fn suc 𝑛 ∧ ( ( 𝑓 ‘ ∅ ) = 𝑥 ∧ ( 𝑓 ‘ 𝑛 ) = 𝑦 ) ∧ ∀ 𝑚 ∈ 𝑛 ( 𝑓 ‘ 𝑚 ) 𝑅 ( 𝑓 ‘ suc 𝑚 ) ) } = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ∃ 𝑛 ∈ ( ω ∖ 1o ) ∃ 𝑓 ( 𝑓 Fn suc 𝑛 ∧ ( ( 𝑓 ‘ ∅ ) = 𝑥 ∧ ( 𝑓 ‘ 𝑛 ) = 𝑦 ) ∧ ∀ 𝑚 ∈ 𝑛 ( 𝑓 ‘ 𝑚 ) 𝑆 ( 𝑓 ‘ suc 𝑚 ) ) } )
7 df-ttrcl ⊢ t++ 𝑅 = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ∃ 𝑛 ∈ ( ω ∖ 1o ) ∃ 𝑓 ( 𝑓 Fn suc 𝑛 ∧ ( ( 𝑓 ‘ ∅ ) = 𝑥 ∧ ( 𝑓 ‘ 𝑛 ) = 𝑦 ) ∧ ∀ 𝑚 ∈ 𝑛 ( 𝑓 ‘ 𝑚 ) 𝑅 ( 𝑓 ‘ suc 𝑚 ) ) }
8 df-ttrcl ⊢ t++ 𝑆 = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ∃ 𝑛 ∈ ( ω ∖ 1o ) ∃ 𝑓 ( 𝑓 Fn suc 𝑛 ∧ ( ( 𝑓 ‘ ∅ ) = 𝑥 ∧ ( 𝑓 ‘ 𝑛 ) = 𝑦 ) ∧ ∀ 𝑚 ∈ 𝑛 ( 𝑓 ‘ 𝑚 ) 𝑆 ( 𝑓 ‘ suc 𝑚 ) ) }
9 6 7 8 3eqtr4g ⊢ ( 𝑅 = 𝑆 → t++ 𝑅 = t++ 𝑆 )