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Proof step 6 from proof nuni (
previous step
)
unique_al(ax + (a0 × d0)' + az), a0 ∈ ax, ∀ a, d, az. a ∈ ax ∧ unique_al(ax + (a × d)' + az) → a ∈ az ⊦ a = a0, a ∈ ax, a ∈ az, a0 ∈ az
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 3 terms: a0, d0, az
Additional information
Used simplifier rules: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; Used terms: a0, d0, az