Metamath Proof Explorer


Theorem csbwrdg

Description: Class substitution for the symbols of a word. (Contributed by Alexander van der Vekens, 15-Jul-2018)

Ref Expression
Assertion csbwrdg S V S / x Word x = Word S

Proof

Step Hyp Ref Expression
1 df-word Word x = w | l 0 w : 0 ..^ l x
2 1 csbeq2i S / x Word x = S / x w | l 0 w : 0 ..^ l x
3 sbcrex [˙S / x]˙ l 0 w : 0 ..^ l x l 0 [˙S / x]˙ w : 0 ..^ l x
4 sbcfg S V [˙S / x]˙ w : 0 ..^ l x S / x w : S / x 0 ..^ l S / x x
5 csbconstg S V S / x w = w
6 csbconstg S V S / x 0 ..^ l = 0 ..^ l
7 csbvarg S V S / x x = S
8 5 6 7 feq123d S V S / x w : S / x 0 ..^ l S / x x w : 0 ..^ l S
9 4 8 bitrd S V [˙S / x]˙ w : 0 ..^ l x w : 0 ..^ l S
10 9 rexbidv S V l 0 [˙S / x]˙ w : 0 ..^ l x l 0 w : 0 ..^ l S
11 3 10 syl5bb S V [˙S / x]˙ l 0 w : 0 ..^ l x l 0 w : 0 ..^ l S
12 11 abbidv S V w | [˙S / x]˙ l 0 w : 0 ..^ l x = w | l 0 w : 0 ..^ l S
13 csbab S / x w | l 0 w : 0 ..^ l x = w | [˙S / x]˙ l 0 w : 0 ..^ l x
14 df-word Word S = w | l 0 w : 0 ..^ l S
15 12 13 14 3eqtr4g S V S / x w | l 0 w : 0 ..^ l x = Word S
16 2 15 syl5eq S V S / x Word x = Word S