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Metamath Proof Explorer

Theorem mpand

Description: A deduction based on modus ponens. (Contributed by NM, 12-Dec-2004) (Proof shortened by Wolf Lammen, 7-Apr-2013)

Ref Expression
Hypotheses mpand.1 ⊢ ( 𝜑 → 𝜓 )
mpand.2 ⊢ ( 𝜑 → ( ( 𝜓 ∧ 𝜒 ) → 𝜃 ) )
Assertion mpand ( 𝜑 → ( 𝜒 → 𝜃 ) )

Proof

Step Hyp Ref Expression
1 mpand.1 ⊢ ( 𝜑 → 𝜓 )
2 mpand.2 ⊢ ( 𝜑 → ( ( 𝜓 ∧ 𝜒 ) → 𝜃 ) )
3 2 ancomsd ⊢ ( 𝜑 → ( ( 𝜒 ∧ 𝜓 ) → 𝜃 ) )
4 1 3 mpan2d ⊢ ( 𝜑 → ( 𝜒 → 𝜃 ) )