Metamath Proof Explorer


Theorem ax13b

Description: An equivalence between two ways of expressing ax-13 . See the comment for ax-13 . (Contributed by NM, 2-May-2017) (Proof shortened by Wolf Lammen, 26-Feb-2018) (Revised by BJ, 15-Sep-2020)

Ref Expression
Assertion ax13b ( ( ¬ 𝑥 = 𝑦 → ( 𝑦 = 𝑧 → 𝜑 ) ) ↔ ( ¬ 𝑥 = 𝑦 → ( ¬ 𝑥 = 𝑧 → ( 𝑦 = 𝑧 → 𝜑 ) ) ) )

Proof

Step Hyp Ref Expression
1 ax-1 ⊢ ( ( 𝑦 = 𝑧 → 𝜑 ) → ( ¬ 𝑥 = 𝑧 → ( 𝑦 = 𝑧 → 𝜑 ) ) )
2 equeuclr ⊢ ( 𝑦 = 𝑧 → ( 𝑥 = 𝑧 → 𝑥 = 𝑦 ) )
3 2 con3rr3 ⊢ ( ¬ 𝑥 = 𝑦 → ( 𝑦 = 𝑧 → ¬ 𝑥 = 𝑧 ) )
4 3 imim1d ⊢ ( ¬ 𝑥 = 𝑦 → ( ( ¬ 𝑥 = 𝑧 → ( 𝑦 = 𝑧 → 𝜑 ) ) → ( 𝑦 = 𝑧 → ( 𝑦 = 𝑧 → 𝜑 ) ) ) )
5 pm2.43 ⊢ ( ( 𝑦 = 𝑧 → ( 𝑦 = 𝑧 → 𝜑 ) ) → ( 𝑦 = 𝑧 → 𝜑 ) )
6 4 5 syl6 ⊢ ( ¬ 𝑥 = 𝑦 → ( ( ¬ 𝑥 = 𝑧 → ( 𝑦 = 𝑧 → 𝜑 ) ) → ( 𝑦 = 𝑧 → 𝜑 ) ) )
7 1 6 impbid2 ⊢ ( ¬ 𝑥 = 𝑦 → ( ( 𝑦 = 𝑧 → 𝜑 ) ↔ ( ¬ 𝑥 = 𝑧 → ( 𝑦 = 𝑧 → 𝜑 ) ) ) )
8 7 pm5.74i ⊢ ( ( ¬ 𝑥 = 𝑦 → ( 𝑦 = 𝑧 → 𝜑 ) ) ↔ ( ¬ 𝑥 = 𝑦 → ( ¬ 𝑥 = 𝑧 → ( 𝑦 = 𝑧 → 𝜑 ) ) ) )