Theorem embantd 56

Description: Deduction embedding an antecedent. (Contributed by Wolf Lammen, 4-Oct-2013.)

Hypotheses

Ref	Expression
embantd.1	⊢ (𝜑 → 𝜓)
embantd.2	⊢ (𝜑 → (𝜒 → 𝜃))

Assertion

Ref	Expression
embantd	⊢ (𝜑 → ((𝜓 → 𝜒) → 𝜃))

Proof of Theorem embantd

Step	Hyp	Ref	Expression
1		embantd.1	. 2 ⊢ (𝜑 → 𝜓)
2		embantd.2	. . 3 ⊢ (𝜑 → (𝜒 → 𝜃))
3	2	imim2d 54	. 2 ⊢ (𝜑 → ((𝜓 → 𝜒) → (𝜓 → 𝜃)))
4	1, 3	mpid 42	1 ⊢ (𝜑 → ((𝜓 → 𝜒) → 𝜃))

This proof depends on syntax axioms:   → wi 4

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

This theorem is used by:  a2and  564  el  4315  findcard2d  7195  findcard2sd  7196  exprmfct  12918  sqrt2irr  12942  pockthg  13138  iscnp4  15321  2sqlem6  16251  bj-exlimmp  16809