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Proof step 6 from proof unique-02 (previous step)

unique_al(ax0), ∀ a1, d2, d3. a1 × d2 ∈ ax0 ∧ a1 × d3 ∈ ax0 → d3 = d2 ⊦ d1 = d0, a ∈ ax0, ¬ a0 × d0 ∈ ax0 ∧ (a0 ≠ a ∨ d0 ≠ d), ¬ a0 × d1 ∈ ax0 ∧ (a0 ≠ a ∨ d1 ≠ d)

User applied rule case distinction

And obtained the following new goals

Proof step 7:
unique_al(ax0), ∀ a1, d2, d3. a1 × d2 ∈ ax0 ∧ a1 × d3 ∈ ax0 → d3 = d2 ⊦ ¬ a0 × d1 ∈ ax0, d1 = d0, a ∈ ax0, ¬ a0 × d0 ∈ ax0 ∧ (a0 ≠ a ∨ d0 ≠ d)

Rule argument

pos: right 4

Additional information

Used simplifier rules: in-01: ⊦ a × d ∈ ax → a ∈ ax;