Description: Alternate proof of dtru which requires more axioms but is shorter and may be easier to understand.
Assuming that ZF set theory is consistent, we cannot prove this theorem unless we specify that x and y be distinct. Specifically, theorem spcev requires that x must not occur in the subexpression -. y = { (/) } in step 4 nor in the subexpression -. y = (/) in step 9. The proof verifier will require that x and y be in a distinct variable group to ensure this. You can check this by deleting the $d statement in set.mm and rerunning the verifier, which will print a detailed explanation of the distinct variable violation. (Contributed by NM, 15-Jul-1994) (Proof modification is discouraged.) (New usage is discouraged.)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | dtruALT | ⊢ ¬ ∀ 𝑥 𝑥 = 𝑦 |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | 0inp0 | ⊢ ( 𝑦 = ∅ → ¬ 𝑦 = { ∅ } ) | |
| 2 | p0ex | ⊢ { ∅ } ∈ V | |
| 3 | eqeq2 | ⊢ ( 𝑥 = { ∅ } → ( 𝑦 = 𝑥 ↔ 𝑦 = { ∅ } ) ) | |
| 4 | 3 | notbid | ⊢ ( 𝑥 = { ∅ } → ( ¬ 𝑦 = 𝑥 ↔ ¬ 𝑦 = { ∅ } ) ) |
| 5 | 2 4 | spcev | ⊢ ( ¬ 𝑦 = { ∅ } → ∃ 𝑥 ¬ 𝑦 = 𝑥 ) |
| 6 | 1 5 | syl | ⊢ ( 𝑦 = ∅ → ∃ 𝑥 ¬ 𝑦 = 𝑥 ) |
| 7 | 0ex | ⊢ ∅ ∈ V | |
| 8 | eqeq2 | ⊢ ( 𝑥 = ∅ → ( 𝑦 = 𝑥 ↔ 𝑦 = ∅ ) ) | |
| 9 | 8 | notbid | ⊢ ( 𝑥 = ∅ → ( ¬ 𝑦 = 𝑥 ↔ ¬ 𝑦 = ∅ ) ) |
| 10 | 7 9 | spcev | ⊢ ( ¬ 𝑦 = ∅ → ∃ 𝑥 ¬ 𝑦 = 𝑥 ) |
| 11 | 6 10 | pm2.61i | ⊢ ∃ 𝑥 ¬ 𝑦 = 𝑥 |
| 12 | exnal | ⊢ ( ∃ 𝑥 ¬ 𝑦 = 𝑥 ↔ ¬ ∀ 𝑥 𝑦 = 𝑥 ) | |
| 13 | eqcom | ⊢ ( 𝑦 = 𝑥 ↔ 𝑥 = 𝑦 ) | |
| 14 | 13 | albii | ⊢ ( ∀ 𝑥 𝑦 = 𝑥 ↔ ∀ 𝑥 𝑥 = 𝑦 ) |
| 15 | 12 14 | xchbinx | ⊢ ( ∃ 𝑥 ¬ 𝑦 = 𝑥 ↔ ¬ ∀ 𝑥 𝑥 = 𝑦 ) |
| 16 | 11 15 | mpbi | ⊢ ¬ ∀ 𝑥 𝑥 = 𝑦 |