Metamath Proof Explorer


Theorem nohalf

Description: An explicit expression for one half. This theorem avoids the axiom of infinity. (Contributed by Scott Fenton, 23-Jul-2025)

Ref Expression
Assertion nohalf ( 1s /su 2s ) = ( { 0s } |s { 1s } )

Proof

Step Hyp Ref Expression
1 twocut ⊢ ( 2s ·s ( { 0s } |s { 1s } ) ) = 1s
2 1no ⊢ 1s ∈ No
3 2 a1i ⊢ ( ⊤ → 1s ∈ No )
4 0no ⊢ 0s ∈ No
5 4 a1i ⊢ ( ⊤ → 0s ∈ No )
6 0lt1s ⊢ 0s <s 1s
7 6 a1i ⊢ ( ⊤ → 0s <s 1s )
8 5 3 7 sltssn ⊢ ( ⊤ → { 0s } <<s { 1s } )
9 8 cutscld ⊢ ( ⊤ → ( { 0s } |s { 1s } ) ∈ No )
10 2no ⊢ 2s ∈ No
11 10 a1i ⊢ ( ⊤ → 2s ∈ No )
12 2ne0s ⊢ 2s ≠ 0s
13 12 a1i ⊢ ( ⊤ → 2s ≠ 0s )
14 oveq2 ⊢ ( 𝑥 = ( { 0s } |s { 1s } ) → ( 2s ·s 𝑥 ) = ( 2s ·s ( { 0s } |s { 1s } ) ) )
15 14 eqeq1d ⊢ ( 𝑥 = ( { 0s } |s { 1s } ) → ( ( 2s ·s 𝑥 ) = 1s ↔ ( 2s ·s ( { 0s } |s { 1s } ) ) = 1s ) )
16 1 a1i ⊢ ( ⊤ → ( 2s ·s ( { 0s } |s { 1s } ) ) = 1s )
17 15 9 16 rspcedvdw ⊢ ( ⊤ → ∃ 𝑥 ∈ No ( 2s ·s 𝑥 ) = 1s )
18 3 9 11 13 17 divmulswd ⊢ ( ⊤ → ( ( 1s /su 2s ) = ( { 0s } |s { 1s } ) ↔ ( 2s ·s ( { 0s } |s { 1s } ) ) = 1s ) )
19 1 18 mpbiri ⊢ ( ⊤ → ( 1s /su 2s ) = ( { 0s } |s { 1s } ) )
20 19 mptru ⊢ ( 1s /su 2s ) = ( { 0s } |s { 1s } )