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Metamath Proof Explorer

Theorem pm5.21ndd

Description: Eliminate an antecedent implied by each side of a biconditional, deduction version. (Contributed by Paul Chapman, 21-Nov-2012) (Proof shortened by Wolf Lammen, 6-Oct-2013)

	Ref	Expression
Hypotheses	pm5.21ndd.1	⊢ ( 𝜑 → ( 𝜒 → 𝜓 ) )
	pm5.21ndd.2	⊢ ( 𝜑 → ( 𝜃 → 𝜓 ) )
	pm5.21ndd.3	⊢ ( 𝜑 → ( 𝜓 → ( 𝜒 ↔ 𝜃 ) ) )
Assertion	pm5.21ndd	⊢ ( 𝜑 → ( 𝜒 ↔ 𝜃 ) )

Proof

Step	Hyp	Ref	Expression
1		pm5.21ndd.1	⊢ ( 𝜑 → ( 𝜒 → 𝜓 ) )
2		pm5.21ndd.2	⊢ ( 𝜑 → ( 𝜃 → 𝜓 ) )
3		pm5.21ndd.3	⊢ ( 𝜑 → ( 𝜓 → ( 𝜒 ↔ 𝜃 ) ) )
4	1	con3d	⊢ ( 𝜑 → ( ¬ 𝜓 → ¬ 𝜒 ) )
5	2	con3d	⊢ ( 𝜑 → ( ¬ 𝜓 → ¬ 𝜃 ) )
6		pm5.21im	⊢ ( ¬ 𝜒 → ( ¬ 𝜃 → ( 𝜒 ↔ 𝜃 ) ) )
7	4 5 6	syl6c	⊢ ( 𝜑 → ( ¬ 𝜓 → ( 𝜒 ↔ 𝜃 ) ) )
8	3 7	pm2.61d	⊢ ( 𝜑 → ( 𝜒 ↔ 𝜃 ) )