Metamath Proof Explorer


Theorem onasuc

Description: Addition with successor. Theorem 4I(A2) of Enderton p. 79. (Note that this version of oasuc does not need Replacement.) (Contributed by Mario Carneiro, 16-Nov-2014)

Ref Expression
Assertion onasuc ⊢ A ∈ On ∧ B ∈ ω → A + 𝑜 suc ⁡ B = suc ⁡ A + 𝑜 B

Proof

Step Hyp Ref Expression
1 frsuc ⊢ B ∈ ω → rec ⁡ x ∈ V ⟼ suc ⁡ x A ↾ ω ⁡ suc ⁡ B = x ∈ V ⟼ suc ⁡ x ⁡ rec ⁡ x ∈ V ⟼ suc ⁡ x A ↾ ω ⁡ B
2 1 adantl ⊢ A ∈ On ∧ B ∈ ω → rec ⁡ x ∈ V ⟼ suc ⁡ x A ↾ ω ⁡ suc ⁡ B = x ∈ V ⟼ suc ⁡ x ⁡ rec ⁡ x ∈ V ⟼ suc ⁡ x A ↾ ω ⁡ B
3 peano2 ⊢ B ∈ ω → suc ⁡ B ∈ ω
4 3 adantl ⊢ A ∈ On ∧ B ∈ ω → suc ⁡ B ∈ ω
5 4 fvresd ⊢ A ∈ On ∧ B ∈ ω → rec ⁡ x ∈ V ⟼ suc ⁡ x A ↾ ω ⁡ suc ⁡ B = rec ⁡ x ∈ V ⟼ suc ⁡ x A ⁡ suc ⁡ B
6 fvres ⊢ B ∈ ω → rec ⁡ x ∈ V ⟼ suc ⁡ x A ↾ ω ⁡ B = rec ⁡ x ∈ V ⟼ suc ⁡ x A ⁡ B
7 6 adantl ⊢ A ∈ On ∧ B ∈ ω → rec ⁡ x ∈ V ⟼ suc ⁡ x A ↾ ω ⁡ B = rec ⁡ x ∈ V ⟼ suc ⁡ x A ⁡ B
8 7 fveq2d ⊢ A ∈ On ∧ B ∈ ω → x ∈ V ⟼ suc ⁡ x ⁡ rec ⁡ x ∈ V ⟼ suc ⁡ x A ↾ ω ⁡ B = x ∈ V ⟼ suc ⁡ x ⁡ rec ⁡ x ∈ V ⟼ suc ⁡ x A ⁡ B
9 2 5 8 3eqtr3d ⊢ A ∈ On ∧ B ∈ ω → rec ⁡ x ∈ V ⟼ suc ⁡ x A ⁡ suc ⁡ B = x ∈ V ⟼ suc ⁡ x ⁡ rec ⁡ x ∈ V ⟼ suc ⁡ x A ⁡ B
10 nnon ⊢ B ∈ ω → B ∈ On
11 onsuc ⊢ B ∈ On → suc ⁡ B ∈ On
12 10 11 syl ⊢ B ∈ ω → suc ⁡ B ∈ On
13 oav ⊢ A ∈ On ∧ suc ⁡ B ∈ On → A + 𝑜 suc ⁡ B = rec ⁡ x ∈ V ⟼ suc ⁡ x A ⁡ suc ⁡ B
14 12 13 sylan2 ⊢ A ∈ On ∧ B ∈ ω → A + 𝑜 suc ⁡ B = rec ⁡ x ∈ V ⟼ suc ⁡ x A ⁡ suc ⁡ B
15 ovex ⊢ A + 𝑜 B ∈ V
16 suceq ⊢ x = A + 𝑜 B → suc ⁡ x = suc ⁡ A + 𝑜 B
17 eqid ⊢ x ∈ V ⟼ suc ⁡ x = x ∈ V ⟼ suc ⁡ x
18 15 sucex ⊢ suc ⁡ A + 𝑜 B ∈ V
19 16 17 18 fvmpt ⊢ A + 𝑜 B ∈ V → x ∈ V ⟼ suc ⁡ x ⁡ A + 𝑜 B = suc ⁡ A + 𝑜 B
20 15 19 ax-mp ⊢ x ∈ V ⟼ suc ⁡ x ⁡ A + 𝑜 B = suc ⁡ A + 𝑜 B
21 oav ⊢ A ∈ On ∧ B ∈ On → A + 𝑜 B = rec ⁡ x ∈ V ⟼ suc ⁡ x A ⁡ B
22 10 21 sylan2 ⊢ A ∈ On ∧ B ∈ ω → A + 𝑜 B = rec ⁡ x ∈ V ⟼ suc ⁡ x A ⁡ B
23 22 fveq2d ⊢ A ∈ On ∧ B ∈ ω → x ∈ V ⟼ suc ⁡ x ⁡ A + 𝑜 B = x ∈ V ⟼ suc ⁡ x ⁡ rec ⁡ x ∈ V ⟼ suc ⁡ x A ⁡ B
24 20 23 eqtr3id ⊢ A ∈ On ∧ B ∈ ω → suc ⁡ A + 𝑜 B = x ∈ V ⟼ suc ⁡ x ⁡ rec ⁡ x ∈ V ⟼ suc ⁡ x A ⁡ B
25 9 14 24 3eqtr4d ⊢ A ∈ On ∧ B ∈ ω → A + 𝑜 suc ⁡ B = suc ⁡ A + 𝑜 B