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Proof step 3 from proof lookup-02 (previous step)

∀ a, d. unique_al(ax0) ∧ a × d ∈ ax0 → ax0[a] = d ⊦ unique_al(p + ax0) ∧ a × d ∈ p + ax0 → (p + ax0)[a] = d

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
unique_al(p ' + ax0), a × d ∈ ax0, ∀ a, d. unique_al(ax0) ∧ a × d ∈ ax0 → ax0[a] = d ⊦ (p ' + ax0)[a] = d, a × d = p

Rule argument

Additional information

Used simplifier rules: assoc-yes: ⊦ ((a × d)' + ax)[a] = d;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;% ⊦ p ∈ p0 ' ↔ p = p0;% ⊦ p ∈ ax + ax0 ↔ ¬ ¬(p ∈ ax ∨ p ∈ ax0);% ⊦ p + ax = p ' + ax;