Theorem prodeq1d 14666

Description: Conditions for two composites to be equal.

Hypothesis

Ref	Expression
prodeq1d.1	|- (ph -> A = B)

Assertion

Ref	Expression
prodeq1d	|- (ph -> prod_k e. AGC = prod_k e. BGC)

Proof of Theorem prodeq1d

Step	Hyp	Ref	Expression
1		prodeq1d.1	. 2 |- (ph -> A = B)
2		prodeq1 14658	. 2 |- (A = B -> prod_k e. AGC = prod_k e. BGC)
3	1, 2	syl 12	1 |- (ph -> prod_k e. AGC = prod_k e. BGC)

Syntax hints:   -> wi 3   = wceq 1298  prod_cprd2 14654

This theorem is referenced by:  fprod1slem 14676  fprodp1slem 14681  clfsebs 14707  fincmpzer 14711  fprodadd 14713  fprodneg 14741

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 1304  ax-gen 1305  ax-8 1306  ax-10 1308  ax-12 1310  ax-17 1317  ax-4 1319  ax-5o 1321  ax-6o 1324  ax-9o 1481  ax-10o 1500  ax-16 1580  ax-11o 1588  ax-ext 1865

This theorem depends on definitions:  df-bi 164  df-or 241  df-an 242  df-ex 1327  df-sb 1536  df-clab 1872  df-cleq 1877  df-clel 1880  df-rex 2110  df-if 2983  df-prod 14653  df-prod2 14655

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