Theorem coeq1d 2506

Description: Equality deduction for composition of two classes.

Hypothesis

Ref	Expression
coeq1d.1	⊢ (φ → A = B)

Assertion

Ref	Expression
coeq1d	⊢ (φ → (A ∘ C) = (B ∘ C))

Proof of Theorem coeq1d

Step	Hyp	Ref	Expression
1		coeq1d.1	. 2 ⊢ (φ → A = B)
2		coeq1 2502	. 2 ⊢ (A = B → (A ∘ C) = (B ∘ C))
3	1, 2	syl 12	1 ⊢ (φ → (A ∘ C) = (B ∘ C))

Syntax hints:   → wi 2   = wceq 1091   ∘ ccom 2414

This theorem is referenced by:  mapenlem1 3384  mapenlem2 3385  seqval 4665

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-br 2063  df-opab 2098  df-co 2427

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