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Proof step 15 from proof last-elim (previous step)

∀ a2, y0. a2 = (a1 ' + x).last ∧ y0 = (a1 ' + x).butlast ↔ a1 ' + x = y0 + a2 ' ⊦ a1 ' + x = (a1 ' + x).butlast + (a1 ' + x).last '

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: (a1 ' + x).last, (a1 ' + x).butlast

Additional information

Used terms: (a1 ' + x).last, (a1 ' + x).butlast