Metamath Proof Explorer


Theorem oprpiece1res1

Description: Restriction to the first part of a piecewise defined function. (Contributed by Jeff Madsen, 11-Jun-2010) (Proof shortened by Mario Carneiro, 3-Sep-2015)

Ref Expression
Hypotheses oprpiece1.1 A
oprpiece1.2 B
oprpiece1.3 A B
oprpiece1.4 R V
oprpiece1.5 S V
oprpiece1.6 K A B
oprpiece1.7 F = x A B , y C if x K R S
oprpiece1.8 G = x A K , y C R
Assertion oprpiece1res1 F A K × C = G

Proof

Step Hyp Ref Expression
1 oprpiece1.1 A
2 oprpiece1.2 B
3 oprpiece1.3 A B
4 oprpiece1.4 R V
5 oprpiece1.5 S V
6 oprpiece1.6 K A B
7 oprpiece1.7 F = x A B , y C if x K R S
8 oprpiece1.8 G = x A K , y C R
9 1 rexri A *
10 2 rexri B *
11 lbicc2 A * B * A B A A B
12 9 10 3 11 mp3an A A B
13 iccss2 A A B K A B A K A B
14 12 6 13 mp2an A K A B
15 ssid C C
16 resmpo A K A B C C x A B , y C if x K R S A K × C = x A K , y C if x K R S
17 14 15 16 mp2an x A B , y C if x K R S A K × C = x A K , y C if x K R S
18 7 reseq1i F A K × C = x A B , y C if x K R S A K × C
19 eliccxr K A B K *
20 6 19 ax-mp K *
21 iccleub A * K * x A K x K
22 9 20 21 mp3an12 x A K x K
23 22 iftrued x A K if x K R S = R
24 23 adantr x A K y C if x K R S = R
25 24 mpoeq3ia x A K , y C if x K R S = x A K , y C R
26 8 25 eqtr4i G = x A K , y C if x K R S
27 17 18 26 3eqtr4i F A K × C = G