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Proof step 7 from proof lookup-02 (previous step)

⊦ unique_al(ax0) ∧ unique_al((a × d)' + ax0) ∧ a0 × d0 ∈ ax0 ∧ (∀ a0, d0. a0 × d0 ∈ ax0 → ax0[a0] = d0) → ((a × d)' + ax0)[a0] = d0 ∨ a0 × d0 = a × d

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: cons: ⊦ unique_al((a × d)' + ax) ↔ ¬ a ∈ ax ∧ unique_al(ax);%unique-02: ⊦ unique_al(ax) ∧ a × d ∈ ax ∧ a × d0 ∈ ax → d = d0;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;