Metamath Proof Explorer


Theorem on2ind

Description: Double induction over ordinal numbers. (Contributed by Scott Fenton, 26-Aug-2024)

Ref Expression
Hypotheses on2ind.1 a = c φ ψ
on2ind.2 b = d ψ χ
on2ind.3 a = c θ χ
on2ind.4 a = X φ τ
on2ind.5 b = Y τ η
on2ind.i a On b On c a d b χ c a ψ d b θ φ
Assertion on2ind X On Y On η

Proof

Step Hyp Ref Expression
1 on2ind.1 a = c φ ψ
2 on2ind.2 b = d ψ χ
3 on2ind.3 a = c θ χ
4 on2ind.4 a = X φ τ
5 on2ind.5 b = Y τ η
6 on2ind.i a On b On c a d b χ c a ψ d b θ φ
7 onfr E Fr On
8 epweon E We On
9 weso E We On E Or On
10 sopo E Or On E Po On
11 8 9 10 mp2b E Po On
12 epse E Se On
13 predon a On Pred E On a = a
14 13 adantr a On b On Pred E On a = a
15 predon b On Pred E On b = b
16 15 adantl a On b On Pred E On b = b
17 16 raleqdv a On b On d Pred E On b χ d b χ
18 14 17 raleqbidv a On b On c Pred E On a d Pred E On b χ c a d b χ
19 14 raleqdv a On b On c Pred E On a ψ c a ψ
20 16 raleqdv a On b On d Pred E On b θ d b θ
21 18 19 20 3anbi123d a On b On c Pred E On a d Pred E On b χ c Pred E On a ψ d Pred E On b θ c a d b χ c a ψ d b θ
22 21 6 sylbid a On b On c Pred E On a d Pred E On b χ c Pred E On a ψ d Pred E On b θ φ
23 7 11 12 7 11 12 1 2 3 4 5 22 xpord2ind X On Y On η