Theorem ancomstVD 45606

Description: Closed form of ancoms 464. The following user's proof is completed by invoking mmj2's unify command and using mmj2's StepSelector to pick all remaining steps of the Metamath proof.

1::	⊢ ((𝜑 ∧ 𝜓) ↔ (𝜓 ∧ 𝜑))
qed:1,?: e0a 45513	⊢ (((𝜑 ∧ 𝜓) → 𝜒) ↔ ((𝜓 ∧ 𝜑) → 𝜒))

The proof of ancomst 470 is derived automatically from it. (Contributed by Alan Sare, 25-Dec-2011.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
ancomstVD	⊢ (((𝜑 ∧ 𝜓) → 𝜒) ↔ ((𝜓 ∧ 𝜑) → 𝜒))

Proof of Theorem ancomstVD

Step	Hyp	Ref	Expression
1		ancom 466	. 2 ⊢ ((𝜑 ∧ 𝜓) ↔ (𝜓 ∧ 𝜑))
2		imbi1 350	. 2 ⊢ (((𝜑 ∧ 𝜓) ↔ (𝜓 ∧ 𝜑)) → (((𝜑 ∧ 𝜓) → 𝜒) ↔ ((𝜓 ∧ 𝜑) → 𝜒)))
3	1, 2	e0a 45513	1 ⊢ (((𝜑 ∧ 𝜓) → 𝜒) ↔ ((𝜓 ∧ 𝜑) → 𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-an 402

This theorem is used by: (None)