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Proof step 1 from proof inkeys-n (conclusion)

⊦ a ≠ b → (a ∈ (b × d0)' + ax ↔ a ∈ ax)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a × d ∈ ax ⊦ a = b, ∃ d1. a × d1 = b × d0 ∨ a × d1 ∈ ax

Rule argument

Additional information

Used simplifier rules: inassoc: ⊦ a ∈ ax ↔ (∃ d. a × d ∈ ax);%in-01: ⊦ a × d ∈ ax → a ∈ ax;% ⊦ ¬ p ∈ p0 ' + ax ↔ p ≠ p0 ∧ ¬ p ∈ ax;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;% ⊦ a × d = a0 × d0 ↔ a = a0 ∧ d = d0;% ⊦ p ∈ p0 ' ↔ p = p0;% ⊦ p ∈ ax + ax0 ↔ ¬ ¬(p ∈ ax ∨ p ∈ ax0);