Metamath Proof Explorer


Theorem uzindd

Description: Induction on the upper integers that start at M . The first four hypotheses give us the substitution instances we need; the following two are the basis and the induction step, a deduction version. (Contributed by metakunt, 8-Jun-2024)

Ref Expression
Hypotheses uzindd.1 ⊢ j = M → ψ ↔ χ
uzindd.2 ⊢ j = k → ψ ↔ θ
uzindd.3 ⊢ j = k + 1 → ψ ↔ τ
uzindd.4 ⊢ j = N → ψ ↔ η
uzindd.5 ⊢ φ → χ
uzindd.6 ⊢ φ ∧ θ ∧ k ∈ ℤ ∧ M ≤ k → τ
uzindd.7 ⊢ φ → M ∈ ℤ
uzindd.8 ⊢ φ → N ∈ ℤ
uzindd.9 ⊢ φ → M ≤ N
Assertion uzindd ⊢ φ → η

Proof

Step Hyp Ref Expression
1 uzindd.1 ⊢ j = M → ψ ↔ χ
2 uzindd.2 ⊢ j = k → ψ ↔ θ
3 uzindd.3 ⊢ j = k + 1 → ψ ↔ τ
4 uzindd.4 ⊢ j = N → ψ ↔ η
5 uzindd.5 ⊢ φ → χ
6 uzindd.6 ⊢ φ ∧ θ ∧ k ∈ ℤ ∧ M ≤ k → τ
7 uzindd.7 ⊢ φ → M ∈ ℤ
8 uzindd.8 ⊢ φ → N ∈ ℤ
9 uzindd.9 ⊢ φ → M ≤ N
10 7 8 9 3jca ⊢ φ → M ∈ ℤ ∧ N ∈ ℤ ∧ M ≤ N
11 1 imbi2d ⊢ j = M → φ → ψ ↔ φ → χ
12 2 imbi2d ⊢ j = k → φ → ψ ↔ φ → θ
13 3 imbi2d ⊢ j = k + 1 → φ → ψ ↔ φ → τ
14 4 imbi2d ⊢ j = N → φ → ψ ↔ φ → η
15 5 adantr ⊢ φ ∧ M ∈ ℤ → χ
16 15 expcom ⊢ M ∈ ℤ → φ → χ
17 3anass ⊢ M ∈ ℤ ∧ k ∈ ℤ ∧ M ≤ k ↔ M ∈ ℤ ∧ k ∈ ℤ ∧ M ≤ k
18 ancom ⊢ M ∈ ℤ ∧ k ∈ ℤ ∧ M ≤ k ↔ k ∈ ℤ ∧ M ≤ k ∧ M ∈ ℤ
19 17 18 bitri ⊢ M ∈ ℤ ∧ k ∈ ℤ ∧ M ≤ k ↔ k ∈ ℤ ∧ M ≤ k ∧ M ∈ ℤ
20 6 ad4ant123 ⊢ φ ∧ θ ∧ k ∈ ℤ ∧ M ≤ k ∧ M ∈ ℤ → τ
21 20 anasss ⊢ φ ∧ θ ∧ k ∈ ℤ ∧ M ≤ k ∧ M ∈ ℤ → τ
22 19 21 sylan2b ⊢ φ ∧ θ ∧ M ∈ ℤ ∧ k ∈ ℤ ∧ M ≤ k → τ
23 22 3impa ⊢ φ ∧ θ ∧ M ∈ ℤ ∧ k ∈ ℤ ∧ M ≤ k → τ
24 23 3com23 ⊢ φ ∧ M ∈ ℤ ∧ k ∈ ℤ ∧ M ≤ k ∧ θ → τ
25 24 3expia ⊢ φ ∧ M ∈ ℤ ∧ k ∈ ℤ ∧ M ≤ k → θ → τ
26 25 expcom ⊢ M ∈ ℤ ∧ k ∈ ℤ ∧ M ≤ k → φ → θ → τ
27 26 a2d ⊢ M ∈ ℤ ∧ k ∈ ℤ ∧ M ≤ k → φ → θ → φ → τ
28 11 12 13 14 16 27 uzind ⊢ M ∈ ℤ ∧ N ∈ ℤ ∧ M ≤ N → φ → η
29 10 28 mpcom ⊢ φ → η