Metamath Proof Explorer


Theorem nn1suc

Description: If a statement holds for 1 and also holds for a successor, it holds for all positive integers. The first three hypotheses give us the substitution instances we need; the last two show that it holds for 1 and for a successor. (Contributed by NM, 11-Oct-2004) (Revised by Mario Carneiro, 16-May-2014)

Ref Expression
Hypotheses nn1suc.1 x=1φψ
nn1suc.3 x=y+1φχ
nn1suc.4 x=Aφθ
nn1suc.5 ψ
nn1suc.6 yχ
Assertion nn1suc Aθ

Proof

Step Hyp Ref Expression
1 nn1suc.1 x=1φψ
2 nn1suc.3 x=y+1φχ
3 nn1suc.4 x=Aφθ
4 nn1suc.5 ψ
5 nn1suc.6 yχ
6 1ex 1V
7 6 1 sbcie [˙1/x]˙φψ
8 4 7 mpbir [˙1/x]˙φ
9 1nn 1
10 eleq1 A=1A1
11 9 10 mpbiri A=1A
12 3 sbcieg A[˙A/x]˙φθ
13 11 12 syl A=1[˙A/x]˙φθ
14 dfsbcq A=1[˙A/x]˙φ[˙1/x]˙φ
15 13 14 bitr3d A=1θ[˙1/x]˙φ
16 8 15 mpbiri A=1θ
17 16 a1i AA=1θ
18 ovex y+1V
19 18 2 sbcie [˙y+1/x]˙φχ
20 oveq1 y=A1y+1=A-1+1
21 20 sbceq1d y=A1[˙y+1/x]˙φ[˙A-1+1/x]˙φ
22 19 21 bitr3id y=A1χ[˙A-1+1/x]˙φ
23 22 5 vtoclga A1[˙A-1+1/x]˙φ
24 nncn AA
25 ax-1cn 1
26 npcan A1A-1+1=A
27 24 25 26 sylancl AA-1+1=A
28 27 sbceq1d A[˙A-1+1/x]˙φ[˙A/x]˙φ
29 28 12 bitrd A[˙A-1+1/x]˙φθ
30 23 29 imbitrid AA1θ
31 nn1m1nn AA=1A1
32 17 30 31 mpjaod Aθ