Theorem ancomstVD 44863

Description: Closed form of ancoms 458. 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 44770	⊢ (((𝜑 ∧ 𝜓) → 𝜒) ↔ ((𝜓 ∧ 𝜑) → 𝜒))

The proof of ancomst 464 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 460	. 2 ⊢ ((𝜑 ∧ 𝜓) ↔ (𝜓 ∧ 𝜑))
2		imbi1 347	. 2 ⊢ (((𝜑 ∧ 𝜓) ↔ (𝜓 ∧ 𝜑)) → (((𝜑 ∧ 𝜓) → 𝜒) ↔ ((𝜓 ∧ 𝜑) → 𝜒)))
3	1, 2	e0a 44770	1 ⊢ (((𝜑 ∧ 𝜓) → 𝜒) ↔ ((𝜓 ∧ 𝜑) → 𝜒))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 207  df-an 396

This theorem is referenced by: (None)