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Proof step 1 from proof firstoc (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