Theorem tapeq2 7620

Description: Equality theorem for tight apartness predicate. (Contributed by Jim Kingdon, 15-Feb-2025.)

Assertion

Ref	Expression
tapeq2	⊢ (𝐴 = 𝐵 → (𝑅 TAp 𝐴 ↔ 𝑅 TAp 𝐵))

Proof of Theorem tapeq2

Dummy variables 𝑥 𝑦 𝑧 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		xpeq12 4793	. . . . 5 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 = 𝐵) → (𝐴 × 𝐴) = (𝐵 × 𝐵))
2	1	anidms 401	. . . 4 ⊢ (𝐴 = 𝐵 → (𝐴 × 𝐴) = (𝐵 × 𝐵))
3	2	sseq2d 3278	. . 3 ⊢ (𝐴 = 𝐵 → (𝑅 ⊆ (𝐴 × 𝐴) ↔ 𝑅 ⊆ (𝐵 × 𝐵)))
4		raleq 2749	. . . 4 ⊢ (𝐴 = 𝐵 → (∀𝑥 ∈ 𝐴 ¬ 𝑥𝑅𝑥 ↔ ∀𝑥 ∈ 𝐵 ¬ 𝑥𝑅𝑥))
5		raleq 2749	. . . . 5 ⊢ (𝐴 = 𝐵 → (∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 → 𝑦𝑅𝑥) ↔ ∀𝑦 ∈ 𝐵 (𝑥𝑅𝑦 → 𝑦𝑅𝑥)))
6	5	raleqbi1dv 2761	. . . 4 ⊢ (𝐴 = 𝐵 → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 → 𝑦𝑅𝑥) ↔ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 (𝑥𝑅𝑦 → 𝑦𝑅𝑥)))
7	4, 6	anbi12d 477	. . 3 ⊢ (𝐴 = 𝐵 → ((∀𝑥 ∈ 𝐴 ¬ 𝑥𝑅𝑥 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 → 𝑦𝑅𝑥)) ↔ (∀𝑥 ∈ 𝐵 ¬ 𝑥𝑅𝑥 ∧ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 (𝑥𝑅𝑦 → 𝑦𝑅𝑥))))
8		raleq 2749	. . . . . 6 ⊢ (𝐴 = 𝐵 → (∀𝑧 ∈ 𝐴 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧)) ↔ ∀𝑧 ∈ 𝐵 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧))))
9	8	raleqbi1dv 2761	. . . . 5 ⊢ (𝐴 = 𝐵 → (∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧)) ↔ ∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧))))
10	9	raleqbi1dv 2761	. . . 4 ⊢ (𝐴 = 𝐵 → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧)) ↔ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧))))
11		raleq 2749	. . . . 5 ⊢ (𝐴 = 𝐵 → (∀𝑦 ∈ 𝐴 (¬ 𝑥𝑅𝑦 → 𝑥 = 𝑦) ↔ ∀𝑦 ∈ 𝐵 (¬ 𝑥𝑅𝑦 → 𝑥 = 𝑦)))
12	11	raleqbi1dv 2761	. . . 4 ⊢ (𝐴 = 𝐵 → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (¬ 𝑥𝑅𝑦 → 𝑥 = 𝑦) ↔ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 (¬ 𝑥𝑅𝑦 → 𝑥 = 𝑦)))
13	10, 12	anbi12d 477	. . 3 ⊢ (𝐴 = 𝐵 → ((∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧)) ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (¬ 𝑥𝑅𝑦 → 𝑥 = 𝑦)) ↔ (∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧)) ∧ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 (¬ 𝑥𝑅𝑦 → 𝑥 = 𝑦))))
14	3, 7, 13	3anbi123d 1353	. 2 ⊢ (𝐴 = 𝐵 → ((𝑅 ⊆ (𝐴 × 𝐴) ∧ (∀𝑥 ∈ 𝐴 ¬ 𝑥𝑅𝑥 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 → 𝑦𝑅𝑥)) ∧ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧)) ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (¬ 𝑥𝑅𝑦 → 𝑥 = 𝑦))) ↔ (𝑅 ⊆ (𝐵 × 𝐵) ∧ (∀𝑥 ∈ 𝐵 ¬ 𝑥𝑅𝑥 ∧ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 (𝑥𝑅𝑦 → 𝑦𝑅𝑥)) ∧ (∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧)) ∧ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 (¬ 𝑥𝑅𝑦 → 𝑥 = 𝑦)))))
15		dftap2 7618	. 2 ⊢ (𝑅 TAp 𝐴 ↔ (𝑅 ⊆ (𝐴 × 𝐴) ∧ (∀𝑥 ∈ 𝐴 ¬ 𝑥𝑅𝑥 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 → 𝑦𝑅𝑥)) ∧ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧)) ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (¬ 𝑥𝑅𝑦 → 𝑥 = 𝑦))))
16		dftap2 7618	. 2 ⊢ (𝑅 TAp 𝐵 ↔ (𝑅 ⊆ (𝐵 × 𝐵) ∧ (∀𝑥 ∈ 𝐵 ¬ 𝑥𝑅𝑥 ∧ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 (𝑥𝑅𝑦 → 𝑦𝑅𝑥)) ∧ (∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧)) ∧ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 (¬ 𝑥𝑅𝑦 → 𝑥 = 𝑦))))
17	14, 15, 16	3bitr4g 223	1 ⊢ (𝐴 = 𝐵 → (𝑅 TAp 𝐴 ↔ 𝑅 TAp 𝐵))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 104   ↔ wb 105   ∨ wo 720   ∧ w3a 1009   = wceq 1402  ∀wral 2528   ⊆ wss 3220   class class class wbr 4130   × cxp 4772   TAp wtap 7615

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-3an 1011  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-ral 2533  df-in 3226  df-ss 3233  df-opab 4193  df-xp 4780  df-pap 7609  df-tap 7616

This theorem is used by:  exmidmotap  7628  isdrngtap  14690  opprdrng  14704