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Metamath Proof Explorer

Theorem bnj155

Description: Technical lemma for bnj153 . This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011) (New usage is discouraged.)

Ref Expression
Hypotheses bnj155.1 ⊢ ( 𝜓1 ↔ [ 𝑔 / 𝑓 ] 𝜓′ )
bnj155.2 ⊢ ( 𝜓′ ↔ ∀ 𝑖 ∈ ω ( suc 𝑖 ∈ 1o → ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) )
Assertion bnj155 ( 𝜓1 ↔ ∀ 𝑖 ∈ ω ( suc 𝑖 ∈ 1o → ( 𝑔 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑔 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) )

Proof

Step Hyp Ref Expression
1 bnj155.1 ⊢ ( 𝜓1 ↔ [ 𝑔 / 𝑓 ] 𝜓′ )
2 bnj155.2 ⊢ ( 𝜓′ ↔ ∀ 𝑖 ∈ ω ( suc 𝑖 ∈ 1o → ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) )
3 2 sbcbii ⊢ ( [ 𝑔 / 𝑓 ] 𝜓′ ↔ [ 𝑔 / 𝑓 ] ∀ 𝑖 ∈ ω ( suc 𝑖 ∈ 1o → ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) )
4 vex ⊢ 𝑔 ∈ V
5 fveq1 ⊢ ( 𝑓 = 𝑔 → ( 𝑓 ‘ suc 𝑖 ) = ( 𝑔 ‘ suc 𝑖 ) )
6 fveq1 ⊢ ( 𝑓 = 𝑔 → ( 𝑓 ‘ 𝑖 ) = ( 𝑔 ‘ 𝑖 ) )
7 6 iuneq1d ⊢ ( 𝑓 = 𝑔 → ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) = ∪ 𝑦 ∈ ( 𝑔 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) )
8 5 7 eqeq12d ⊢ ( 𝑓 = 𝑔 → ( ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ↔ ( 𝑔 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑔 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) )
9 8 imbi2d ⊢ ( 𝑓 = 𝑔 → ( ( suc 𝑖 ∈ 1o → ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) ↔ ( suc 𝑖 ∈ 1o → ( 𝑔 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑔 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) ) )
10 9 ralbidv ⊢ ( 𝑓 = 𝑔 → ( ∀ 𝑖 ∈ ω ( suc 𝑖 ∈ 1o → ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) ↔ ∀ 𝑖 ∈ ω ( suc 𝑖 ∈ 1o → ( 𝑔 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑔 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) ) )
11 4 10 sbcie ⊢ ( [ 𝑔 / 𝑓 ] ∀ 𝑖 ∈ ω ( suc 𝑖 ∈ 1o → ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) ↔ ∀ 𝑖 ∈ ω ( suc 𝑖 ∈ 1o → ( 𝑔 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑔 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) )
12 1 3 11 3bitri ⊢ ( 𝜓1 ↔ ∀ 𝑖 ∈ ω ( suc 𝑖 ∈ 1o → ( 𝑔 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑔 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) )