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Proof step 3 from proof neq-07 (previous step)

y1 + a0 ' + y0 + a ' = y2 + a ' + a1 ' + z0 ⊦ a = a1, a = a0, a1 = a0, a1 ∈ y0, a ∈ y1

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
⊦ x ≠ [] → (∃ y, a. x = y + a ')
Proof step 5:
y1 + a0 ' + y0 + a ' = y2 + a ' + a1 ' + z0, z0 ≠ [] → (∃ y, a. z0 = y + a ') ⊦ a = a0, a = a1, a1 = a0, a ∈ y1, a1 ∈ y0

Rule argument

spec: <toplevel> name: lastdiv seq: ⊦ x ≠ [] → (∃ y, a. x = y + a ') subst: [x] → [z0]

Additional information

Used terms: z0