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Proof step 1 from proof firstoccurrence-key (conclusion)

⊦ a ∈ ax ↔ (∃ ax0, d, ax1. ax = ax0 + ((a × d) + ax1) ∧ ¬ a ∈ ax0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ∈ ax ⊦ ∃ ax0, d, ax1. ax = ax0 + (a × d)' + ax1 ∧ ¬ a ∈ ax0

Rule argument

Additional information

Used simplifier rules: in-append: ⊦ a ∈ ax + ay ↔ a ∈ ax ∨ a ∈ ay;%inone: ⊦ a ∈ (b × d0)' ↔ a = b;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;% ⊦ p + ax = p ' + ax;