Metamath Proof Explorer


Theorem sucneqond

Description: Inequality of an ordinal set with its successor. Does not use the axiom of regularity. (Contributed by ML, 18-Oct-2020)

Ref Expression
Hypotheses sucneqond.1 ( 𝜑𝑋 = suc 𝑌 )
sucneqond.2 ( 𝜑𝑌 ∈ On )
Assertion sucneqond ( 𝜑𝑋𝑌 )

Proof

Step Hyp Ref Expression
1 sucneqond.1 ( 𝜑𝑋 = suc 𝑌 )
2 sucneqond.2 ( 𝜑𝑌 ∈ On )
3 sucidg ( 𝑌 ∈ On → 𝑌 ∈ suc 𝑌 )
4 2 3 syl ( 𝜑𝑌 ∈ suc 𝑌 )
5 4 1 eleqtrrd ( 𝜑𝑌𝑋 )
6 suceloni ( 𝑌 ∈ On → suc 𝑌 ∈ On )
7 2 6 syl ( 𝜑 → suc 𝑌 ∈ On )
8 1 7 eqeltrd ( 𝜑𝑋 ∈ On )
9 eloni ( 𝑋 ∈ On → Ord 𝑋 )
10 8 9 syl ( 𝜑 → Ord 𝑋 )
11 ordirr ( Ord 𝑋 → ¬ 𝑋𝑋 )
12 10 11 syl ( 𝜑 → ¬ 𝑋𝑋 )
13 eleq1 ( 𝑋 = 𝑌 → ( 𝑋𝑋𝑌𝑋 ) )
14 13 biimprd ( 𝑋 = 𝑌 → ( 𝑌𝑋𝑋𝑋 ) )
15 14 con3d ( 𝑋 = 𝑌 → ( ¬ 𝑋𝑋 → ¬ 𝑌𝑋 ) )
16 12 15 syl5com ( 𝜑 → ( 𝑋 = 𝑌 → ¬ 𝑌𝑋 ) )
17 5 16 mt2d ( 𝜑 → ¬ 𝑋 = 𝑌 )
18 17 neqned ( 𝜑𝑋𝑌 )