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Proof step 5 from proof nuni (previous step)

⊦ unique_al(((a × d)' + ax) + (a0 × d0)' + az) ∧ (∀ a, d, az. a ∈ ax ∧ unique_al(ax + (a × d)' + az) → a ∈ az) → a0 ∈ az ∨ ¬ a0 ∈ (a × d)' ∧ ¬ a0 ∈ ax

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
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

Rule argument

Additional information

Used simplifier rules: cons: ⊦ unique_al((a × d)' + ax) ↔ ¬ a ∈ ax ∧ unique_al(ax);%inapp: ⊦ a ∈ ax + ay ↔ ¬ ¬(a ∈ ax ∨ a ∈ ay);%inone: ⊦ a ∈ (b × d0)' ↔ a = b;%nuni-01: ⊦ unique_al(ay + (a × d)' + az) → ¬ a ∈ az;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;