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

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

User applied rule case distinction

And obtained the following new goals

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

Rule argument

pos: left 2

Additional information

Used simplifier rules: in-01: ⊦ a × d ∈ ax → a ∈ ax;%inj: ⊦ a × d = a0 × d0 ↔ a = a0 ∧ d = d0; from specification pair