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Proof step 8 from proof last-elim (previous step)

y = a3 + y1, ∀ a2, y0. a2 = (a1 ' + x).last ∧ y0 = (a1 ' + x).butlast ↔ a1 ' + x = y0 + a2 ' ⊦ a = (a1 ' + x).last ∧ y = a0 ' + (a1 ' + x).butlast ↔ a0 ' + a1 ' + x = y + a '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∀ a2, y0. a2 = (a1 ' + x).last ∧ y0 = (a1 ' + x).butlast ↔ a1 ' + x = y0 + a2 ' ⊦ a = (a1 ' + x).last ∧ a3 ' + y1 = a0 ' + (a1 ' + x).butlast ↔ a0 ' + a1 ' + x = (a3 ' + y1) + a '

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)