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Proof step 5 from proof nuni (previous step)

unique_al(((a0 × d0)' + ay0) + (a × d)' + az), ∀ a, d, az. a ∈ ay0 ∧ unique_al(ay0 + (a × d)' + az) → a ∈ az, a ∈ (a0 × d0)' ∨ a ∈ ay0 ⊦ a ∈ az

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
unique_al(ay0 + (a × d)' + az), a ∈ ay0, ∀ a, d, az. a ∈ ay0 ∧ unique_al(ay0 + (a × d)' + az) → a ∈ az ⊦ a ∈ az, a0 ∈ az, a0 ∈ ay0

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;%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[assoclista]