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Theorem nebidc 2404
 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 2365 . . 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 2399 . . . . 5 ((𝐴𝐵𝐶𝐷) → (DECID 𝐴 = 𝐵 → (DECID 𝐶 = 𝐷 → (𝐴 = 𝐵𝐶 = 𝐷))))
76com3l 81 . . . 4 (DECID 𝐴 = 𝐵 → (DECID 𝐶 = 𝐷 → ((𝐴𝐵𝐶𝐷) → (𝐴 = 𝐵𝐶 = 𝐷))))
87imp 123 . . 3 ((DECID 𝐴 = 𝐵DECID 𝐶 = 𝐷) → ((𝐴𝐵𝐶𝐷) → (𝐴 = 𝐵𝐶 = 𝐷)))
92, 8impbid2 142 . 2 ((DECID 𝐴 = 𝐵DECID 𝐶 = 𝐷) → ((𝐴 = 𝐵𝐶 = 𝐷) ↔ (𝐴𝐵𝐶𝐷)))
109ex 114 1 (DECID 𝐴 = 𝐵 → (DECID 𝐶 = 𝐷 → ((𝐴 = 𝐵𝐶 = 𝐷) ↔ (𝐴𝐵𝐶𝐷))))
 Colors of variables: wff set class Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104  DECID wdc 820   = wceq 1332   ≠ wne 2324 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 604  ax-in2 605  ax-io 699 This theorem depends on definitions:  df-bi 116  df-stab 817  df-dc 821  df-ne 2325 This theorem is referenced by:  rpexp  11999
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