Theorem 3anandis 1325

Description: Inference that undistributes a triple conjunction in the antecedent. (Contributed by NM, 18-Apr-2007.)

Hypothesis

Ref	Expression
3anandis.1	⊢ (((𝜑 ∧ 𝜓) ∧ (𝜑 ∧ 𝜒) ∧ (𝜑 ∧ 𝜃)) → 𝜏)

Assertion

Ref	Expression
3anandis	⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒 ∧ 𝜃)) → 𝜏)

Proof of Theorem 3anandis

Step	Hyp	Ref	Expression
1		simpl 108	. 2 ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒 ∧ 𝜃)) → 𝜑)
2		simpr1 987	. 2 ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒 ∧ 𝜃)) → 𝜓)
3		simpr2 988	. 2 ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒 ∧ 𝜃)) → 𝜒)
4		simpr3 989	. 2 ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒 ∧ 𝜃)) → 𝜃)
5		3anandis.1	. 2 ⊢ (((𝜑 ∧ 𝜓) ∧ (𝜑 ∧ 𝜒) ∧ (𝜑 ∧ 𝜃)) → 𝜏)
6	1, 2, 1, 3, 1, 4, 5	syl222anc 1232	1 ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒 ∧ 𝜃)) → 𝜏)

Syntax hints:   → wi 4   ∧ wa 103   ∧ w3a 962

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107

This theorem depends on definitions:  df-bi 116  df-3an 964

This theorem is referenced by: (None)