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Metamath Proof Explorer

Theorem bnj953

Description: Technical lemma for bnj69 . 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 bnj953.1 ⊢ ( 𝜓 ↔ ∀ 𝑖 ∈ ω ( suc 𝑖 ∈ 𝑛 → ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) )
bnj953.2 ⊢ ( ( 𝐺 ‘ 𝑖 ) = ( 𝑓 ‘ 𝑖 ) → ∀ 𝑦 ( 𝐺 ‘ 𝑖 ) = ( 𝑓 ‘ 𝑖 ) )
Assertion bnj953 ( ( ( 𝐺 ‘ 𝑖 ) = ( 𝑓 ‘ 𝑖 ) ∧ ( 𝐺 ‘ suc 𝑖 ) = ( 𝑓 ‘ suc 𝑖 ) ∧ ( 𝑖 ∈ ω ∧ suc 𝑖 ∈ 𝑛 ) ∧ 𝜓 ) → ( 𝐺 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝐺 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) )

Proof

Step Hyp Ref Expression
1 bnj953.1 ⊢ ( 𝜓 ↔ ∀ 𝑖 ∈ ω ( suc 𝑖 ∈ 𝑛 → ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) )
2 bnj953.2 ⊢ ( ( 𝐺 ‘ 𝑖 ) = ( 𝑓 ‘ 𝑖 ) → ∀ 𝑦 ( 𝐺 ‘ 𝑖 ) = ( 𝑓 ‘ 𝑖 ) )
3 bnj312 ⊢ ( ( ( 𝐺 ‘ 𝑖 ) = ( 𝑓 ‘ 𝑖 ) ∧ ( 𝐺 ‘ suc 𝑖 ) = ( 𝑓 ‘ suc 𝑖 ) ∧ ( 𝑖 ∈ ω ∧ suc 𝑖 ∈ 𝑛 ) ∧ 𝜓 ) ↔ ( ( 𝐺 ‘ suc 𝑖 ) = ( 𝑓 ‘ suc 𝑖 ) ∧ ( 𝐺 ‘ 𝑖 ) = ( 𝑓 ‘ 𝑖 ) ∧ ( 𝑖 ∈ ω ∧ suc 𝑖 ∈ 𝑛 ) ∧ 𝜓 ) )
4 bnj251 ⊢ ( ( ( 𝐺 ‘ suc 𝑖 ) = ( 𝑓 ‘ suc 𝑖 ) ∧ ( 𝐺 ‘ 𝑖 ) = ( 𝑓 ‘ 𝑖 ) ∧ ( 𝑖 ∈ ω ∧ suc 𝑖 ∈ 𝑛 ) ∧ 𝜓 ) ↔ ( ( 𝐺 ‘ suc 𝑖 ) = ( 𝑓 ‘ suc 𝑖 ) ∧ ( ( 𝐺 ‘ 𝑖 ) = ( 𝑓 ‘ 𝑖 ) ∧ ( ( 𝑖 ∈ ω ∧ suc 𝑖 ∈ 𝑛 ) ∧ 𝜓 ) ) ) )
5 3 4 bitri ⊢ ( ( ( 𝐺 ‘ 𝑖 ) = ( 𝑓 ‘ 𝑖 ) ∧ ( 𝐺 ‘ suc 𝑖 ) = ( 𝑓 ‘ suc 𝑖 ) ∧ ( 𝑖 ∈ ω ∧ suc 𝑖 ∈ 𝑛 ) ∧ 𝜓 ) ↔ ( ( 𝐺 ‘ suc 𝑖 ) = ( 𝑓 ‘ suc 𝑖 ) ∧ ( ( 𝐺 ‘ 𝑖 ) = ( 𝑓 ‘ 𝑖 ) ∧ ( ( 𝑖 ∈ ω ∧ suc 𝑖 ∈ 𝑛 ) ∧ 𝜓 ) ) ) )
6 1 bnj115 ⊢ ( 𝜓 ↔ ∀ 𝑖 ( ( 𝑖 ∈ ω ∧ suc 𝑖 ∈ 𝑛 ) → ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) )
7 sp ⊢ ( ∀ 𝑖 ( ( 𝑖 ∈ ω ∧ suc 𝑖 ∈ 𝑛 ) → ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) → ( ( 𝑖 ∈ ω ∧ suc 𝑖 ∈ 𝑛 ) → ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) )
8 7 impcom ⊢ ( ( ( 𝑖 ∈ ω ∧ suc 𝑖 ∈ 𝑛 ) ∧ ∀ 𝑖 ( ( 𝑖 ∈ ω ∧ suc 𝑖 ∈ 𝑛 ) → ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) ) → ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) )
9 6 8 sylan2b ⊢ ( ( ( 𝑖 ∈ ω ∧ suc 𝑖 ∈ 𝑛 ) ∧ 𝜓 ) → ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) )
10 2 bnj956 ⊢ ( ( 𝐺 ‘ 𝑖 ) = ( 𝑓 ‘ 𝑖 ) → ∪ 𝑦 ∈ ( 𝐺 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) )
11 eqtr3 ⊢ ( ( ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ∧ ∪ 𝑦 ∈ ( 𝐺 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) → ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝐺 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) )
12 9 10 11 syl2anr ⊢ ( ( ( 𝐺 ‘ 𝑖 ) = ( 𝑓 ‘ 𝑖 ) ∧ ( ( 𝑖 ∈ ω ∧ suc 𝑖 ∈ 𝑛 ) ∧ 𝜓 ) ) → ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝐺 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) )
13 eqtr ⊢ ( ( ( 𝐺 ‘ suc 𝑖 ) = ( 𝑓 ‘ suc 𝑖 ) ∧ ( 𝑓 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝐺 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) ) → ( 𝐺 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝐺 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) )
14 12 13 sylan2 ⊢ ( ( ( 𝐺 ‘ suc 𝑖 ) = ( 𝑓 ‘ suc 𝑖 ) ∧ ( ( 𝐺 ‘ 𝑖 ) = ( 𝑓 ‘ 𝑖 ) ∧ ( ( 𝑖 ∈ ω ∧ suc 𝑖 ∈ 𝑛 ) ∧ 𝜓 ) ) ) → ( 𝐺 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝐺 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) )
15 5 14 sylbi ⊢ ( ( ( 𝐺 ‘ 𝑖 ) = ( 𝑓 ‘ 𝑖 ) ∧ ( 𝐺 ‘ suc 𝑖 ) = ( 𝑓 ‘ suc 𝑖 ) ∧ ( 𝑖 ∈ ω ∧ suc 𝑖 ∈ 𝑛 ) ∧ 𝜓 ) → ( 𝐺 ‘ suc 𝑖 ) = ∪ 𝑦 ∈ ( 𝐺 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) )