Metamath Proof Explorer


Theorem frrlem16

Description: Lemma for general founded recursion. Establish a subset relationship. (Contributed by Scott Fenton, 11-Sep-2023)

Ref Expression
Assertion frrlem16 ( ( ( 𝑅 Fr 𝐴𝑅 Se 𝐴 ) ∧ 𝑧𝐴 ) → ∀ 𝑤 ∈ TrPred ( 𝑅 , 𝐴 , 𝑧 ) Pred ( 𝑅 , 𝐴 , 𝑤 ) ⊆ TrPred ( 𝑅 , 𝐴 , 𝑧 ) )

Proof

Step Hyp Ref Expression
1 simplr ( ( ( ( 𝑅 Fr 𝐴𝑅 Se 𝐴 ) ∧ 𝑧𝐴 ) ∧ 𝑤 ∈ TrPred ( 𝑅 , 𝐴 , 𝑧 ) ) → 𝑧𝐴 )
2 simpllr ( ( ( ( 𝑅 Fr 𝐴𝑅 Se 𝐴 ) ∧ 𝑧𝐴 ) ∧ 𝑤 ∈ TrPred ( 𝑅 , 𝐴 , 𝑧 ) ) → 𝑅 Se 𝐴 )
3 1 2 jca ( ( ( ( 𝑅 Fr 𝐴𝑅 Se 𝐴 ) ∧ 𝑧𝐴 ) ∧ 𝑤 ∈ TrPred ( 𝑅 , 𝐴 , 𝑧 ) ) → ( 𝑧𝐴𝑅 Se 𝐴 ) )
4 simpr ( ( ( ( 𝑅 Fr 𝐴𝑅 Se 𝐴 ) ∧ 𝑧𝐴 ) ∧ 𝑤 ∈ TrPred ( 𝑅 , 𝐴 , 𝑧 ) ) → 𝑤 ∈ TrPred ( 𝑅 , 𝐴 , 𝑧 ) )
5 trpredtr ( ( 𝑧𝐴𝑅 Se 𝐴 ) → ( 𝑤 ∈ TrPred ( 𝑅 , 𝐴 , 𝑧 ) → Pred ( 𝑅 , 𝐴 , 𝑤 ) ⊆ TrPred ( 𝑅 , 𝐴 , 𝑧 ) ) )
6 3 4 5 sylc ( ( ( ( 𝑅 Fr 𝐴𝑅 Se 𝐴 ) ∧ 𝑧𝐴 ) ∧ 𝑤 ∈ TrPred ( 𝑅 , 𝐴 , 𝑧 ) ) → Pred ( 𝑅 , 𝐴 , 𝑤 ) ⊆ TrPred ( 𝑅 , 𝐴 , 𝑧 ) )
7 6 ralrimiva ( ( ( 𝑅 Fr 𝐴𝑅 Se 𝐴 ) ∧ 𝑧𝐴 ) → ∀ 𝑤 ∈ TrPred ( 𝑅 , 𝐴 , 𝑧 ) Pred ( 𝑅 , 𝐴 , 𝑤 ) ⊆ TrPred ( 𝑅 , 𝐴 , 𝑧 ) )