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Proof step 3 from proof unique-add (previous step)

⊦ unique_al(ax + (a × d)') ↔ ¬ a ∈ ax ∧ unique_al(ax)

User applied rule case distinction

And obtained the following new goals

Proof step 4:
¬ a ∈ ax ∧ unique_al(ax) ⊦ unique_al(ax + (a × d)')

Rule argument

pos: right 1

Additional information

Used simplifier rules: unique-01: ⊦ unique_al(ax + (a × d)') → ¬ a ∈ ax;%unique: ⊦ unique_al(ax + p ') → unique_al(ax);% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;