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Theorem nebidc 2335
Description: Contraposition law for inequality. (Contributed by Jim Kingdon, 19-May-2018.)
Assertion
Ref Expression
nebidc (DECID 𝐴 = 𝐵 → (DECID 𝐶 = 𝐷 → ((𝐴 = 𝐵𝐶 = 𝐷) ↔ (𝐴𝐵𝐶𝐷))))

Proof of Theorem nebidc
StepHypRef Expression
1 id 19 . . . 4 ((𝐴 = 𝐵𝐶 = 𝐷) → (𝐴 = 𝐵𝐶 = 𝐷))
21necon3bid 2296 . . 3 ((𝐴 = 𝐵𝐶 = 𝐷) → (𝐴𝐵𝐶𝐷))
3 id 19 . . . . . . . 8 ((𝐴𝐵𝐶𝐷) → (𝐴𝐵𝐶𝐷))
43a1d 22 . . . . . . 7 ((𝐴𝐵𝐶𝐷) → (DECID 𝐶 = 𝐷 → (𝐴𝐵𝐶𝐷)))
54a1d 22 . . . . . 6 ((𝐴𝐵𝐶𝐷) → (DECID 𝐴 = 𝐵 → (DECID 𝐶 = 𝐷 → (𝐴𝐵𝐶𝐷))))
65necon4biddc 2330 . . . . 5 ((𝐴𝐵𝐶𝐷) → (DECID 𝐴 = 𝐵 → (DECID 𝐶 = 𝐷 → (𝐴 = 𝐵𝐶 = 𝐷))))
76com3l 80 . . . 4 (DECID 𝐴 = 𝐵 → (DECID 𝐶 = 𝐷 → ((𝐴𝐵𝐶𝐷) → (𝐴 = 𝐵𝐶 = 𝐷))))
87imp 122 . . 3 ((DECID 𝐴 = 𝐵DECID 𝐶 = 𝐷) → ((𝐴𝐵𝐶𝐷) → (𝐴 = 𝐵𝐶 = 𝐷)))
92, 8impbid2 141 . 2 ((DECID 𝐴 = 𝐵DECID 𝐶 = 𝐷) → ((𝐴 = 𝐵𝐶 = 𝐷) ↔ (𝐴𝐵𝐶𝐷)))
109ex 113 1 (DECID 𝐴 = 𝐵 → (DECID 𝐶 = 𝐷 → ((𝐴 = 𝐵𝐶 = 𝐷) ↔ (𝐴𝐵𝐶𝐷))))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 102  wb 103  DECID wdc 780   = wceq 1289  wne 2255
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 579  ax-in2 580  ax-io 665
This theorem depends on definitions:  df-bi 115  df-dc 781  df-ne 2256
This theorem is referenced by:  rpexp  11214
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