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Metamath Proof Explorer

Theorem div2subd

Description: Swap subtrahend and minuend inside the numerator and denominator of a fraction. Deduction form of div2sub . (Contributed by David Moews, 28-Feb-2017)

		Ref	Expression
	Hypotheses	div2subd.1	⊢ ( 𝜑 → 𝐴 ∈ ℂ )
		div2subd.2	⊢ ( 𝜑 → 𝐵 ∈ ℂ )
		div2subd.3	⊢ ( 𝜑 → 𝐶 ∈ ℂ )
		div2subd.4	⊢ ( 𝜑 → 𝐷 ∈ ℂ )
		div2subd.5	⊢ ( 𝜑 → 𝐶 ≠ 𝐷 )
	Assertion	div2subd	⊢ ( 𝜑 → ( ( 𝐴 − 𝐵 ) / ( 𝐶 − 𝐷 ) ) = ( ( 𝐵 − 𝐴 ) / ( 𝐷 − 𝐶 ) ) )

Proof

Step	Hyp	Ref	Expression
1		div2subd.1	⊢ ( 𝜑 → 𝐴 ∈ ℂ )
2		div2subd.2	⊢ ( 𝜑 → 𝐵 ∈ ℂ )
3		div2subd.3	⊢ ( 𝜑 → 𝐶 ∈ ℂ )
4		div2subd.4	⊢ ( 𝜑 → 𝐷 ∈ ℂ )
5		div2subd.5	⊢ ( 𝜑 → 𝐶 ≠ 𝐷 )
6		div2sub	⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) ∧ ( 𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ ∧ 𝐶 ≠ 𝐷 ) ) → ( ( 𝐴 − 𝐵 ) / ( 𝐶 − 𝐷 ) ) = ( ( 𝐵 − 𝐴 ) / ( 𝐷 − 𝐶 ) ) )
7	1 2 3 4 5 6	syl23anc	⊢ ( 𝜑 → ( ( 𝐴 − 𝐵 ) / ( 𝐶 − 𝐷 ) ) = ( ( 𝐵 − 𝐴 ) / ( 𝐷 − 𝐶 ) ) )