Theorem supeq1d 13595

Description: Equality deduction for supremum.

Hypothesis

Ref	Expression
supeq1d.1	|- (ph -> B = C)

Assertion

Ref	Expression
supeq1d	|- (ph -> sup(B, A, R) = sup(C, A, R))

Proof of Theorem supeq1d

Step	Hyp	Ref	Expression
1		supeq1d.1	. 2 |- (ph -> B = C)
2		supeq1 5665	. 2 |- (B = C -> sup(B, A, R) = sup(C, A, R))
3	1, 2	syl 12	1 |- (ph -> sup(B, A, R) = sup(C, A, R))

Syntax hints:   -> wi 3   = wceq 1298  supcsup 5663

This theorem is referenced by:  gcdval 13715

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-9 1307  ax-10 1308  ax-11 1309  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-ral 2109  df-rex 2110  df-rab 2112  df-uni 3178  df-sup 5664

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