Theorem nebidc 2285

Description: Contraposition law for inequality. (Contributed by Jim Kingdon, 19-May-2018.)

Assertion

Ref	Expression
nebidc	⊢ (DECID 𝐴 = 𝐵 → (DECID 𝐶 = 𝐷 → ((𝐴 = 𝐵 ↔ 𝐶 = 𝐷) ↔ (𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷))))

Proof of Theorem nebidc

Step	Hyp	Ref	Expression
1		id 19	. . . 4 ⊢ ((𝐴 = 𝐵 ↔ 𝐶 = 𝐷) → (𝐴 = 𝐵 ↔ 𝐶 = 𝐷))
2	1	necon3bid 2246	. . 3 ⊢ ((𝐴 = 𝐵 ↔ 𝐶 = 𝐷) → (𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷))
3		id 19	. . . . . . . 8 ⊢ ((𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷) → (𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷))
4	3	a1d 22	. . . . . . 7 ⊢ ((𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷) → (DECID 𝐶 = 𝐷 → (𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷)))
5	4	a1d 22	. . . . . 6 ⊢ ((𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷) → (DECID 𝐴 = 𝐵 → (DECID 𝐶 = 𝐷 → (𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷))))
6	5	necon4biddc 2280	. . . . 5 ⊢ ((𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷) → (DECID 𝐴 = 𝐵 → (DECID 𝐶 = 𝐷 → (𝐴 = 𝐵 ↔ 𝐶 = 𝐷))))
7	6	com3l 75	. . . 4 ⊢ (DECID 𝐴 = 𝐵 → (DECID 𝐶 = 𝐷 → ((𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷) → (𝐴 = 𝐵 ↔ 𝐶 = 𝐷))))
8	7	imp 115	. . 3 ⊢ ((DECID 𝐴 = 𝐵 ∧ DECID 𝐶 = 𝐷) → ((𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷) → (𝐴 = 𝐵 ↔ 𝐶 = 𝐷)))
9	2, 8	impbid2 131	. 2 ⊢ ((DECID 𝐴 = 𝐵 ∧ DECID 𝐶 = 𝐷) → ((𝐴 = 𝐵 ↔ 𝐶 = 𝐷) ↔ (𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷)))
10	9	ex 108	1 ⊢ (DECID 𝐴 = 𝐵 → (DECID 𝐶 = 𝐷 → ((𝐴 = 𝐵 ↔ 𝐶 = 𝐷) ↔ (𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷))))

Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98  DECID wdc 742   = wceq 1243   ≠ wne 2204

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 630

This theorem depends on definitions:  df-bi 110  df-dc 743  df-ne 2206

This theorem is referenced by: (None)