Theorem ndmovg 6817

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 6653	. 2 ⊢ (𝐴𝐹𝐵) = (𝐹‘⟨𝐴, 𝐵⟩)
2		eleq2 2690	. . . . . 6 ⊢ (dom 𝐹 = (𝑅 × 𝑆) → (⟨𝐴, 𝐵⟩ ∈ dom 𝐹 ↔ ⟨𝐴, 𝐵⟩ ∈ (𝑅 × 𝑆)))
3		opelxp 5146	. . . . . 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 308	. . . 4 ⊢ (dom 𝐹 = (𝑅 × 𝑆) → (¬ ⟨𝐴, 𝐵⟩ ∈ dom 𝐹 ↔ ¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)))
6		ndmfv 6218	. . . 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 445	. 2 ⊢ ((dom 𝐹 = (𝑅 × 𝑆) ∧ ¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)) → (𝐹‘⟨𝐴, 𝐵⟩) = ∅)
9	1, 8	syl5eq 2668	1 ⊢ ((dom 𝐹 = (𝑅 × 𝑆) ∧ ¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)) → (𝐴𝐹𝐵) = ∅)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 384   = wceq 1483   ∈ wcel 1990  ∅c0 3915  ⟨cop 4183   × cxp 5112  dom cdm 5114  ‘cfv 5888  (class class class)co 6650

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-8 1992  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602  ax-sep 4781  ax-nul 4789  ax-pow 4843  ax-pr 4906

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1039  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-eu 2474  df-mo 2475  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-ral 2917  df-rex 2918  df-rab 2921  df-v 3202  df-dif 3577  df-un 3579  df-in 3581  df-ss 3588  df-nul 3916  df-if 4087  df-sn 4178  df-pr 4180  df-op 4184  df-uni 4437  df-br 4654  df-opab 4713  df-xp 5120  df-dm 5124  df-iota 5851  df-fv 5896  df-ov 6653

This theorem is referenced by:  ndmov  6818  curry1val  7270  curry2val  7274  1div0  10686  repsundef  13518  cshnz  13538  mamufacex  20195  mavmulsolcl  20357  mavmul0g  20359  iscau2  23075  1div0apr  27324