Theorem ndmovg 6984

Description: The value of an operation outside its domain. (Contributed by NM, 28-Mar-2008.)

Assertion

Ref	Expression
ndmovg	⊢ ((dom 𝐹 = (𝑅 × 𝑆) ∧ ¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)) → (𝐴𝐹𝐵) = ∅)

Proof of Theorem ndmovg

Step	Hyp	Ref	Expression
1		df-ov 6818	. 2 ⊢ (𝐴𝐹𝐵) = (𝐹‘⟨𝐴, 𝐵⟩)
2		eleq2 2829	. . . . . 6 ⊢ (dom 𝐹 = (𝑅 × 𝑆) → (⟨𝐴, 𝐵⟩ ∈ dom 𝐹 ↔ ⟨𝐴, 𝐵⟩ ∈ (𝑅 × 𝑆)))
3		opelxp 5304	. . . . . 6 ⊢ (⟨𝐴, 𝐵⟩ ∈ (𝑅 × 𝑆) ↔ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆))
4	2, 3	syl6bb 276	. . . . 5 ⊢ (dom 𝐹 = (𝑅 × 𝑆) → (⟨𝐴, 𝐵⟩ ∈ dom 𝐹 ↔ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)))
5	4	notbid 307	. . . 4 ⊢ (dom 𝐹 = (𝑅 × 𝑆) → (¬ ⟨𝐴, 𝐵⟩ ∈ dom 𝐹 ↔ ¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)))
6		ndmfv 6381	. . . 4 ⊢ (¬ ⟨𝐴, 𝐵⟩ ∈ dom 𝐹 → (𝐹‘⟨𝐴, 𝐵⟩) = ∅)
7	5, 6	syl6bir 244	. . 3 ⊢ (dom 𝐹 = (𝑅 × 𝑆) → (¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆) → (𝐹‘⟨𝐴, 𝐵⟩) = ∅))
8	7	imp 444	. 2 ⊢ ((dom 𝐹 = (𝑅 × 𝑆) ∧ ¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)) → (𝐹‘⟨𝐴, 𝐵⟩) = ∅)
9	1, 8	syl5eq 2807	1 ⊢ ((dom 𝐹 = (𝑅 × 𝑆) ∧ ¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)) → (𝐴𝐹𝐵) = ∅)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 383   = wceq 1632   ∈ wcel 2140  ∅c0 4059  ⟨cop 4328   × cxp 5265  dom cdm 5267  ‘cfv 6050  (class class class)co 6815

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1871  ax-4 1886  ax-5 1989  ax-6 2055  ax-7 2091  ax-8 2142  ax-9 2149  ax-10 2169  ax-11 2184  ax-12 2197  ax-13 2392  ax-ext 2741  ax-sep 4934  ax-nul 4942  ax-pow 4993  ax-pr 5056

This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-3an 1074  df-tru 1635  df-ex 1854  df-nf 1859  df-sb 2048  df-eu 2612  df-mo 2613  df-clab 2748  df-cleq 2754  df-clel 2757  df-nfc 2892  df-ral 3056  df-rex 3057  df-rab 3060  df-v 3343  df-dif 3719  df-un 3721  df-in 3723  df-ss 3730  df-nul 4060  df-if 4232  df-sn 4323  df-pr 4325  df-op 4329  df-uni 4590  df-br 4806  df-opab 4866  df-xp 5273  df-dm 5277  df-iota 6013  df-fv 6058  df-ov 6818

This theorem is referenced by:  ndmov  6985  curry1val  7440  curry2val  7444  1div0  10899  repsundef  13739  cshnz  13759  mamufacex  20418  mavmulsolcl  20580  mavmul0g  20582  iscau2  23296  1div0apr  27657