Metamath Proof Explorer


Theorem oa1suc

Description: Addition with 1 is same as successor. Proposition 4.34(a) of Mendelson p. 266. Remark 2.4 of Schloeder p. 4. (Contributed by NM, 29-Oct-1995) (Revised by Mario Carneiro, 16-Nov-2014)

Ref Expression
Assertion oa1suc ( 𝐴 ∈ On → ( 𝐴 +o 1o ) = suc 𝐴 )

Proof

Step Hyp Ref Expression
1 df-1o ⊢ 1o = suc ∅
2 1 oveq2i ⊢ ( 𝐴 +o 1o ) = ( 𝐴 +o suc ∅ )
3 peano1 ⊢ ∅ ∈ ω
4 onasuc ⊢ ( ( 𝐴 ∈ On ∧ ∅ ∈ ω ) → ( 𝐴 +o suc ∅ ) = suc ( 𝐴 +o ∅ ) )
5 3 4 mpan2 ⊢ ( 𝐴 ∈ On → ( 𝐴 +o suc ∅ ) = suc ( 𝐴 +o ∅ ) )
6 2 5 eqtrid ⊢ ( 𝐴 ∈ On → ( 𝐴 +o 1o ) = suc ( 𝐴 +o ∅ ) )
7 oa0 ⊢ ( 𝐴 ∈ On → ( 𝐴 +o ∅ ) = 𝐴 )
8 suceq ⊢ ( ( 𝐴 +o ∅ ) = 𝐴 → suc ( 𝐴 +o ∅ ) = suc 𝐴 )
9 7 8 syl ⊢ ( 𝐴 ∈ On → suc ( 𝐴 +o ∅ ) = suc 𝐴 )
10 6 9 eqtrd ⊢ ( 𝐴 ∈ On → ( 𝐴 +o 1o ) = suc 𝐴 )