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Theorem nebidc 2279
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 2240 . . 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 2274 . . . . 5 ((AB𝐶𝐷) → (DECID A = B → (DECID 𝐶 = 𝐷 → (A = B𝐶 = 𝐷))))
76com3l 75 . . . 4 (DECID A = B → (DECID 𝐶 = 𝐷 → ((AB𝐶𝐷) → (A = B𝐶 = 𝐷))))
87imp 115 . . 3 ((DECID A = B DECID 𝐶 = 𝐷) → ((AB𝐶𝐷) → (A = B𝐶 = 𝐷)))
92, 8impbid2 131 . 2 ((DECID A = B DECID 𝐶 = 𝐷) → ((A = B𝐶 = 𝐷) ↔ (AB𝐶𝐷)))
109ex 108 1 (DECID A = B → (DECID 𝐶 = 𝐷 → ((A = B𝐶 = 𝐷) ↔ (AB𝐶𝐷))))
Colors of variables: wff set class
Syntax hints:  wi 4   wa 97  wb 98  DECID wdc 741   = wceq 1242  wne 2201
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-io 629
This theorem depends on definitions:  df-bi 110  df-dc 742  df-ne 2203
This theorem is referenced by: (None)
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