Metamath Proof Explorer


Theorem bnj986

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 bnj986.3 χ n D f Fn n φ ψ
bnj986.10 D = ω
bnj986.15 τ m ω n = suc m p = suc n
Assertion bnj986 χ m p τ

Proof

Step Hyp Ref Expression
1 bnj986.3 χ n D f Fn n φ ψ
2 bnj986.10 D = ω
3 bnj986.15 τ m ω n = suc m p = suc n
4 2 bnj158 n D m ω n = suc m
5 1 4 bnj769 χ m ω n = suc m
6 5 bnj1196 χ m m ω n = suc m
7 vex n V
8 7 sucex suc n V
9 8 isseti p p = suc n
10 6 9 jctir χ m m ω n = suc m p p = suc n
11 exdistr m p m ω n = suc m p = suc n m m ω n = suc m p p = suc n
12 19.41v m m ω n = suc m p p = suc n m m ω n = suc m p p = suc n
13 11 12 bitr2i m m ω n = suc m p p = suc n m p m ω n = suc m p = suc n
14 10 13 sylib χ m p m ω n = suc m p = suc n
15 df-3an m ω n = suc m p = suc n m ω n = suc m p = suc n
16 3 15 bitri τ m ω n = suc m p = suc n
17 16 2exbii m p τ m p m ω n = suc m p = suc n
18 14 17 sylibr χ m p τ