We can define a field F with the following properties:

• Binary operations + (addition) and ⋅ (multiplication)
• Commutativity
• Associativity
• Identities
• Inverses
• Distributivity

Now, the additive inverse existence condition can be equivalently replaced with the statement ∀a∈F 0a=0, because each one is a consequence of the other.

The proof (assuming 0=0a): $$0=0a=(1-1)a=a+(-1a)$$ Which means that for every a there exists an additive inverse, more precisely it is -1a.

But now take the set {0,1} with logical conjunction for ⋅; and inclusive disjunction for +. (so that 1+1=1)

• 1 doesn't have an additive inverse, as 1+1=1+0=1≠0
• concurrently, 0⋅0=0 and 0⋅1=0; so this indeed appears to be a field by the alternative condition.