Description: Alternate proof of axc14 (even when inlining the above results, this gives a shorter proof). (Contributed by BJ, 20-Oct-2021) (Proof modification is discouraged.)
Ref | Expression | ||
---|---|---|---|
Assertion | bj-axc14 | ⊢ ( ¬ ∀ 𝑧 𝑧 = 𝑥 → ( ¬ ∀ 𝑧 𝑧 = 𝑦 → ( 𝑥 ∈ 𝑦 → ∀ 𝑧 𝑥 ∈ 𝑦 ) ) ) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | bj-axc14nf | ⊢ ( ¬ ∀ 𝑧 𝑧 = 𝑥 → ( ¬ ∀ 𝑧 𝑧 = 𝑦 → Ⅎ 𝑧 𝑥 ∈ 𝑦 ) ) | |
2 | nf5r | ⊢ ( Ⅎ 𝑧 𝑥 ∈ 𝑦 → ( 𝑥 ∈ 𝑦 → ∀ 𝑧 𝑥 ∈ 𝑦 ) ) | |
3 | 2 | a1i | ⊢ ( ¬ ∀ 𝑧 𝑧 = 𝑥 → ( Ⅎ 𝑧 𝑥 ∈ 𝑦 → ( 𝑥 ∈ 𝑦 → ∀ 𝑧 𝑥 ∈ 𝑦 ) ) ) |
4 | 1 3 | syld | ⊢ ( ¬ ∀ 𝑧 𝑧 = 𝑥 → ( ¬ ∀ 𝑧 𝑧 = 𝑦 → ( 𝑥 ∈ 𝑦 → ∀ 𝑧 𝑥 ∈ 𝑦 ) ) ) |