Metamath Proof Explorer

Theorem bnj548

Description: Technical lemma for bnj852 . 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 bnj548.1 ⊢ ( 𝜏 ↔ ( 𝑓 Fn 𝑚 ∧ 𝜑′ ∧ 𝜓′ ) )
bnj548.2 ⊢ 𝐵 = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 )
bnj548.3 ⊢ 𝐾 = ∪ 𝑦 ∈ ( 𝐺 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 )
bnj548.4 ⊢ 𝐺 = ( 𝑓 ∪ { ⟨ 𝑚 , 𝐶 ⟩ } )
bnj548.5 ⊢ ( ( 𝑅 FrSe 𝐴 ∧ 𝜏 ∧ 𝜎 ) → 𝐺 Fn 𝑛 )
Assertion bnj548 ( ( ( 𝑅 FrSe 𝐴 ∧ 𝜏 ∧ 𝜎 ) ∧ 𝑖 ∈ 𝑚 ) → 𝐵 = 𝐾 )

Proof

Step Hyp Ref Expression
1 bnj548.1 ⊢ ( 𝜏 ↔ ( 𝑓 Fn 𝑚 ∧ 𝜑′ ∧ 𝜓′ ) )
2 bnj548.2 ⊢ 𝐵 = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 )
3 bnj548.3 ⊢ 𝐾 = ∪ 𝑦 ∈ ( 𝐺 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 )
4 bnj548.4 ⊢ 𝐺 = ( 𝑓 ∪ { ⟨ 𝑚 , 𝐶 ⟩ } )
5 bnj548.5 ⊢ ( ( 𝑅 FrSe 𝐴 ∧ 𝜏 ∧ 𝜎 ) → 𝐺 Fn 𝑛 )
6 5 bnj930 ⊢ ( ( 𝑅 FrSe 𝐴 ∧ 𝜏 ∧ 𝜎 ) → Fun 𝐺 )
7 6 adantr ⊢ ( ( ( 𝑅 FrSe 𝐴 ∧ 𝜏 ∧ 𝜎 ) ∧ 𝑖 ∈ 𝑚 ) → Fun 𝐺 )
8 1 simp1bi ⊢ ( 𝜏 → 𝑓 Fn 𝑚 )
9 fndm ⊢ ( 𝑓 Fn 𝑚 → dom 𝑓 = 𝑚 )
10 eleq2 ⊢ ( dom 𝑓 = 𝑚 → ( 𝑖 ∈ dom 𝑓 ↔ 𝑖 ∈ 𝑚 ) )
11 10 biimpar ⊢ ( ( dom 𝑓 = 𝑚 ∧ 𝑖 ∈ 𝑚 ) → 𝑖 ∈ dom 𝑓 )
12 9 11 sylan ⊢ ( ( 𝑓 Fn 𝑚 ∧ 𝑖 ∈ 𝑚 ) → 𝑖 ∈ dom 𝑓 )
13 8 12 sylan ⊢ ( ( 𝜏 ∧ 𝑖 ∈ 𝑚 ) → 𝑖 ∈ dom 𝑓 )
14 13 3ad2antl2 ⊢ ( ( ( 𝑅 FrSe 𝐴 ∧ 𝜏 ∧ 𝜎 ) ∧ 𝑖 ∈ 𝑚 ) → 𝑖 ∈ dom 𝑓 )
15 7 14 jca ⊢ ( ( ( 𝑅 FrSe 𝐴 ∧ 𝜏 ∧ 𝜎 ) ∧ 𝑖 ∈ 𝑚 ) → ( Fun 𝐺 ∧ 𝑖 ∈ dom 𝑓 ) )
16 4 bnj931 ⊢ 𝑓 ⊆ 𝐺
17 15 16 jctil ⊢ ( ( ( 𝑅 FrSe 𝐴 ∧ 𝜏 ∧ 𝜎 ) ∧ 𝑖 ∈ 𝑚 ) → ( 𝑓 ⊆ 𝐺 ∧ ( Fun 𝐺 ∧ 𝑖 ∈ dom 𝑓 ) ) )
18 3anan12 ⊢ ( ( Fun 𝐺 ∧ 𝑓 ⊆ 𝐺 ∧ 𝑖 ∈ dom 𝑓 ) ↔ ( 𝑓 ⊆ 𝐺 ∧ ( Fun 𝐺 ∧ 𝑖 ∈ dom 𝑓 ) ) )
19 17 18 sylibr ⊢ ( ( ( 𝑅 FrSe 𝐴 ∧ 𝜏 ∧ 𝜎 ) ∧ 𝑖 ∈ 𝑚 ) → ( Fun 𝐺 ∧ 𝑓 ⊆ 𝐺 ∧ 𝑖 ∈ dom 𝑓 ) )
20 funssfv ⊢ ( ( Fun 𝐺 ∧ 𝑓 ⊆ 𝐺 ∧ 𝑖 ∈ dom 𝑓 ) → ( 𝐺 ‘ 𝑖 ) = ( 𝑓 ‘ 𝑖 ) )
21 iuneq1 ⊢ ( ( 𝐺 ‘ 𝑖 ) = ( 𝑓 ‘ 𝑖 ) → ∪ 𝑦 ∈ ( 𝐺 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) = ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) )
22 21 eqcomd ⊢ ( ( 𝐺 ‘ 𝑖 ) = ( 𝑓 ‘ 𝑖 ) → ∪ 𝑦 ∈ ( 𝑓 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) = ∪ 𝑦 ∈ ( 𝐺 ‘ 𝑖 ) pred ( 𝑦 , 𝐴 , 𝑅 ) )
23 22 2 3 3eqtr4g ⊢ ( ( 𝐺 ‘ 𝑖 ) = ( 𝑓 ‘ 𝑖 ) → 𝐵 = 𝐾 )
24 19 20 23 3syl ⊢ ( ( ( 𝑅 FrSe 𝐴 ∧ 𝜏 ∧ 𝜎 ) ∧ 𝑖 ∈ 𝑚 ) → 𝐵 = 𝐾 )