Metamath Proof Explorer


Theorem brsuccf

Description: Binary relation form of the Succ function. (Contributed by Scott Fenton, 14-Apr-2014)

Ref Expression
Hypotheses brsuccf.1 ⊢ A ∈ V
brsuccf.2 ⊢ B ∈ V
Assertion brsuccf ⊢ A 𝖲𝗎𝖼𝖼 B ↔ B = suc ⁡ A

Proof

Step Hyp Ref Expression
1 brsuccf.1 ⊢ A ∈ V
2 brsuccf.2 ⊢ B ∈ V
3 df-succf ⊢ 𝖲𝗎𝖼𝖼 = 𝖢𝗎𝗉 ∘ I ⊗ 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇
4 3 breqi ⊢ A 𝖲𝗎𝖼𝖼 B ↔ A 𝖢𝗎𝗉 ∘ I ⊗ 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 B
5 1 2 brco ⊢ A 𝖢𝗎𝗉 ∘ I ⊗ 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 B ↔ ∃ x A I ⊗ 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 x ∧ x 𝖢𝗎𝗉 B
6 1 2 lemsuccf ⊢ ∃ x A I ⊗ 𝖲𝗂𝗇𝗀𝗅𝖾𝗍𝗈𝗇 x ∧ x 𝖢𝗎𝗉 B ↔ B = suc ⁡ A
7 4 5 6 3bitri ⊢ A 𝖲𝗎𝖼𝖼 B ↔ B = suc ⁡ A