Theorem fvreseq1 6898

Description: Equality of a function restricted to the domain of another function. (Contributed by AV, 6-Jan-2019.)

Assertion

Ref	Expression
fvreseq1	⊢ (((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐵) ∧ 𝐵 ⊆ 𝐴) → ((𝐹 ↾ 𝐵) = 𝐺 ↔ ∀𝑥 ∈ 𝐵 (𝐹‘𝑥) = (𝐺‘𝑥)))

Distinct variable groups:   𝑥,𝐵   𝑥,𝐹   𝑥,𝐺

Allowed substitution hint:   𝐴(𝑥)

Proof of Theorem fvreseq1

Step	Hyp	Ref	Expression
1		fnresdm 6535	. . . . 5 ⊢ (𝐺 Fn 𝐵 → (𝐺 ↾ 𝐵) = 𝐺)
2	1	ad2antlr 723	. . . 4 ⊢ (((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐵) ∧ 𝐵 ⊆ 𝐴) → (𝐺 ↾ 𝐵) = 𝐺)
3	2	eqcomd 2744	. . 3 ⊢ (((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐵) ∧ 𝐵 ⊆ 𝐴) → 𝐺 = (𝐺 ↾ 𝐵))
4	3	eqeq2d 2749	. 2 ⊢ (((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐵) ∧ 𝐵 ⊆ 𝐴) → ((𝐹 ↾ 𝐵) = 𝐺 ↔ (𝐹 ↾ 𝐵) = (𝐺 ↾ 𝐵)))
5		ssid 3939	. . 3 ⊢ 𝐵 ⊆ 𝐵
6		fvreseq0 6897	. . 3 ⊢ (((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐵) ∧ (𝐵 ⊆ 𝐴 ∧ 𝐵 ⊆ 𝐵)) → ((𝐹 ↾ 𝐵) = (𝐺 ↾ 𝐵) ↔ ∀𝑥 ∈ 𝐵 (𝐹‘𝑥) = (𝐺‘𝑥)))
7	5, 6	mpanr2 700	. 2 ⊢ (((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐵) ∧ 𝐵 ⊆ 𝐴) → ((𝐹 ↾ 𝐵) = (𝐺 ↾ 𝐵) ↔ ∀𝑥 ∈ 𝐵 (𝐹‘𝑥) = (𝐺‘𝑥)))
8	4, 7	bitrd 278	1 ⊢ (((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐵) ∧ 𝐵 ⊆ 𝐴) → ((𝐹 ↾ 𝐵) = 𝐺 ↔ ∀𝑥 ∈ 𝐵 (𝐹‘𝑥) = (𝐺‘𝑥)))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395   = wceq 1539  ∀wral 3063   ⊆ wss 3883   ↾ cres 5582   Fn wfn 6413  ‘cfv 6418

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-10 2139  ax-11 2156  ax-12 2173  ax-ext 2709  ax-sep 5218  ax-nul 5225  ax-pr 5347

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-3an 1087  df-tru 1542  df-fal 1552  df-ex 1784  df-nf 1788  df-sb 2069  df-mo 2540  df-eu 2569  df-clab 2716  df-cleq 2730  df-clel 2817  df-nfc 2888  df-ne 2943  df-ral 3068  df-rex 3069  df-rab 3072  df-v 3424  df-sbc 3712  df-csb 3829  df-dif 3886  df-un 3888  df-in 3890  df-ss 3900  df-nul 4254  df-if 4457  df-sn 4559  df-pr 4561  df-op 4565  df-uni 4837  df-br 5071  df-opab 5133  df-mpt 5154  df-id 5480  df-xp 5586  df-rel 5587  df-cnv 5588  df-co 5589  df-dm 5590  df-rn 5591  df-res 5592  df-ima 5593  df-iota 6376  df-fun 6420  df-fn 6421  df-fv 6426

This theorem is referenced by:  symgextres  18948  sseqfres  32260