Metamath Proof Explorer


Theorem oacomf1olem

Description: Lemma for oacomf1o . (Contributed by Mario Carneiro, 30-May-2015)

Ref Expression
Hypothesis oacomf1olem.1 F = x A B + 𝑜 x
Assertion oacomf1olem A On B On F : A 1-1 onto ran F ran F B =

Proof

Step Hyp Ref Expression
1 oacomf1olem.1 F = x A B + 𝑜 x
2 oaf1o B On x On B + 𝑜 x : On 1-1 onto On B
3 2 adantl A On B On x On B + 𝑜 x : On 1-1 onto On B
4 f1of1 x On B + 𝑜 x : On 1-1 onto On B x On B + 𝑜 x : On 1-1 On B
5 3 4 syl A On B On x On B + 𝑜 x : On 1-1 On B
6 onss A On A On
7 6 adantr A On B On A On
8 f1ssres x On B + 𝑜 x : On 1-1 On B A On x On B + 𝑜 x A : A 1-1 On B
9 5 7 8 syl2anc A On B On x On B + 𝑜 x A : A 1-1 On B
10 7 resmptd A On B On x On B + 𝑜 x A = x A B + 𝑜 x
11 10 1 eqtr4di A On B On x On B + 𝑜 x A = F
12 f1eq1 x On B + 𝑜 x A = F x On B + 𝑜 x A : A 1-1 On B F : A 1-1 On B
13 11 12 syl A On B On x On B + 𝑜 x A : A 1-1 On B F : A 1-1 On B
14 9 13 mpbid A On B On F : A 1-1 On B
15 f1f1orn F : A 1-1 On B F : A 1-1 onto ran F
16 14 15 syl A On B On F : A 1-1 onto ran F
17 f1f F : A 1-1 On B F : A On B
18 frn F : A On B ran F On B
19 14 17 18 3syl A On B On ran F On B
20 19 difss2d A On B On ran F On
21 reldisj ran F On ran F B = ran F On B
22 20 21 syl A On B On ran F B = ran F On B
23 19 22 mpbird A On B On ran F B =
24 16 23 jca A On B On F : A 1-1 onto ran F ran F B =