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Proof step 11 from proof last-elim (previous step)

∀ 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 = a3 ∧ a1 ' + x = y1 + a '

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ a ' + x = b ' + y ↔ a = b ∧ x = y
Proof step 13:
∀ a2, y0. a2 = (a1 ' + x).last ∧ y0 = (a1 ' + x).butlast ↔ a1 ' + x = y0 + a2 ' ⊦ a = (a1 ' + x).last ∧ a3 = a0 ∧ y1 = (a1 ' + x).butlast ↔ a0 = a3 ∧ a1 ' + x = y1 + a '

Rule argument

spec: list name: c-06 seq: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y subst: [y, b, x, a] → [(a1 ' + x).butlast, a0, y1, a3] flags: <all paths>

Additional information

Used terms: (a1 ' + x).butlast, a0, y1, a3