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Mirrors > Home > ILE Home > Th. List > eqneqall | GIF version |
Description: A contradiction concerning equality implies anything. (Contributed by Alexander van der Vekens, 25-Jan-2018.) |
Ref | Expression |
---|---|
eqneqall | ⊢ (𝐴 = 𝐵 → (𝐴 ≠ 𝐵 → 𝜑)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | df-ne 2335 | . 2 ⊢ (𝐴 ≠ 𝐵 ↔ ¬ 𝐴 = 𝐵) | |
2 | pm2.24 611 | . 2 ⊢ (𝐴 = 𝐵 → (¬ 𝐴 = 𝐵 → 𝜑)) | |
3 | 1, 2 | syl5bi 151 | 1 ⊢ (𝐴 = 𝐵 → (𝐴 ≠ 𝐵 → 𝜑)) |
Colors of variables: wff set class |
Syntax hints: ¬ wn 3 → wi 4 = wceq 1342 ≠ wne 2334 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-in2 605 |
This theorem depends on definitions: df-bi 116 df-ne 2335 |
This theorem is referenced by: eldju2ndl 7029 eldju2ndr 7030 modfzo0difsn 10321 nno 11832 prm2orodd 12047 prm23lt5 12184 dvdsprmpweqnn 12256 logbgcd1irr 13452 |
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