Theorem supeq1d 6400

Description: Equality deduction for supremum. (Contributed by Paul Chapman, 22-Jun-2011.)

Hypothesis

Ref	Expression
supeq1d.1	⊢ (𝜑 → 𝐵 = 𝐶)

Assertion

Ref	Expression
supeq1d	⊢ (𝜑 → sup(𝐵, 𝐴, 𝑅) = sup(𝐶, 𝐴, 𝑅))

Proof of Theorem supeq1d

Step	Hyp	Ref	Expression
1		supeq1d.1	. 2 ⊢ (𝜑 → 𝐵 = 𝐶)
2		supeq1 6399	. 2 ⊢ (𝐵 = 𝐶 → sup(𝐵, 𝐴, 𝑅) = sup(𝐶, 𝐴, 𝑅))
3	1, 2	syl 14	1 ⊢ (𝜑 → sup(𝐵, 𝐴, 𝑅) = sup(𝐶, 𝐴, 𝑅))

Syntax hints:   → wi 4   = wceq 1284  supcsup 6395

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-ral 2353  df-rex 2354  df-rab 2357  df-uni 3602  df-sup 6397

This theorem is referenced by:  supminfex  8685  minmax  10112  gcdval  10351  gcdass  10404