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Metamath Proof Explorer

Theorem infeq1i

Description: Equality inference for infimum. (Contributed by AV, 2-Sep-2020)

		Ref	Expression
	Hypothesis	infeq1i.1	$⊢ B = C$
	Assertion	infeq1i	$⊢ inf (B, A, R) = inf (C, A, R)$

Step	Hyp	Ref	Expression
1		infeq1i.1	$⊢ B = C$
2		infeq1	$⊢ B = C \to inf (B, A, R) = inf (C, A, R)$
3	1 2	ax-mp	$⊢ inf (B, A, R) = inf (C, A, R)$