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