Metamath Proof Explorer


Theorem findes

Description: Finite induction with explicit substitution. The first hypothesis is the basis and the second is the induction step. Theorem Schema 22 of Suppes p. 136. See tfindes for the transfinite version. This is an alternative for Metamath 100 proof #74. (Contributed by Raph Levien, 9-Jul-2003)

Ref Expression
Hypotheses findes.1 [˙/x]˙φ
findes.2 xωφ[˙sucx/x]˙φ
Assertion findes xωφ

Proof

Step Hyp Ref Expression
1 findes.1 [˙/x]˙φ
2 findes.2 xωφ[˙sucx/x]˙φ
3 dfsbcq2 z=zxφ[˙/x]˙φ
4 sbequ z=yzxφyxφ
5 dfsbcq2 z=sucyzxφ[˙sucy/x]˙φ
6 sbequ12r z=xzxφφ
7 nfv xyω
8 nfs1v xyxφ
9 nfsbc1v x[˙sucy/x]˙φ
10 8 9 nfim xyxφ[˙sucy/x]˙φ
11 7 10 nfim xyωyxφ[˙sucy/x]˙φ
12 eleq1w x=yxωyω
13 sbequ12 x=yφyxφ
14 suceq x=ysucx=sucy
15 14 sbceq1d x=y[˙sucx/x]˙φ[˙sucy/x]˙φ
16 13 15 imbi12d x=yφ[˙sucx/x]˙φyxφ[˙sucy/x]˙φ
17 12 16 imbi12d x=yxωφ[˙sucx/x]˙φyωyxφ[˙sucy/x]˙φ
18 11 17 2 chvarfv yωyxφ[˙sucy/x]˙φ
19 3 4 5 6 1 18 finds xωφ