Theorem pm5.21nd 902

Description: Eliminate an antecedent implied by each side of a biconditional. (Contributed by NM, 20-Nov-2005.) (Proof shortened by Wolf Lammen, 4-Nov-2013.)

Hypotheses

Ref	Expression
pm5.21nd.1	⊢ ((𝜑 ∧ 𝜓) → 𝜃)
pm5.21nd.2	⊢ ((𝜑 ∧ 𝜒) → 𝜃)
pm5.21nd.3	⊢ (𝜃 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
pm5.21nd	⊢ (𝜑 → (𝜓 ↔ 𝜒))

Proof of Theorem pm5.21nd

Step	Hyp	Ref	Expression
1		pm5.21nd.1	. . 3 ⊢ ((𝜑 ∧ 𝜓) → 𝜃)
2	1	ex 114	. 2 ⊢ (𝜑 → (𝜓 → 𝜃))
3		pm5.21nd.2	. . 3 ⊢ ((𝜑 ∧ 𝜒) → 𝜃)
4	3	ex 114	. 2 ⊢ (𝜑 → (𝜒 → 𝜃))
5		pm5.21nd.3	. . 3 ⊢ (𝜃 → (𝜓 ↔ 𝜒))
6	5	a1i 9	. 2 ⊢ (𝜑 → (𝜃 → (𝜓 ↔ 𝜒)))
7	2, 4, 6	pm5.21ndd 695	1 ⊢ (𝜑 → (𝜓 ↔ 𝜒))

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104

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

This theorem is referenced by:  ideqg  4698  fvelimab  5485  releldm2  6091  relelec  6477  fzrev3  9898  elfzp12  9910  eltg  12260  eltg2  12261  cncnp2m  12439