Theorem had1 1605

Description: If the first input is true, then the adder sum is equivalent to the biconditionality of the other two inputs. (Contributed by Mario Carneiro, 4-Sep-2016.) (Proof shortened by Wolf Lammen, 11-Jul-2020.)

Assertion

Ref	Expression
had1	⊢ (𝜑 → (hadd(𝜑, 𝜓, 𝜒) ↔ (𝜓 ↔ 𝜒)))

Proof of Theorem had1

Step	Hyp	Ref	Expression
1		hadrot 1603	. . . 4 ⊢ (hadd(𝜑, 𝜓, 𝜒) ↔ hadd(𝜓, 𝜒, 𝜑))
2		hadbi 1599	. . . 4 ⊢ (hadd(𝜓, 𝜒, 𝜑) ↔ ((𝜓 ↔ 𝜒) ↔ 𝜑))
3	1, 2	bitri 274	. . 3 ⊢ (hadd(𝜑, 𝜓, 𝜒) ↔ ((𝜓 ↔ 𝜒) ↔ 𝜑))
4		biass 386	. . 3 ⊢ (((hadd(𝜑, 𝜓, 𝜒) ↔ (𝜓 ↔ 𝜒)) ↔ 𝜑) ↔ (hadd(𝜑, 𝜓, 𝜒) ↔ ((𝜓 ↔ 𝜒) ↔ 𝜑)))
5	3, 4	mpbir 230	. 2 ⊢ ((hadd(𝜑, 𝜓, 𝜒) ↔ (𝜓 ↔ 𝜒)) ↔ 𝜑)
6	5	biimpri 227	1 ⊢ (𝜑 → (hadd(𝜑, 𝜓, 𝜒) ↔ (𝜓 ↔ 𝜒)))

Syntax hints:   → wi 4   ↔ wb 205  haddwhad 1594

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 206  df-xor 1507  df-had 1595

This theorem is referenced by:  had0  1606  hadifp  1607  sadadd2lem2  16157