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Proof step 1 from proof put-one-01 (conclusion)
a ≠ a0 ⊦ (a × d)'[a0, d0] = (a × d)' + (a0 × d0)'
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 2
:
⊦ a = a0, (a × d)'[a0, d0] = (a × d)' + (a0 × d0)'
Rule argument
Additional information