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Proof step 2 from proof ni (previous step)

x + a ' = y + z, a ∈ y ⊦ z = [], a ∈ z

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ x ≠ [] → (∃ y, a. x = y + a ')

Rule argument

spec: list inst: name: lastdiv seq: ⊦ x ≠ [] → (∃ y, a. x = y + a ') subst: [x] → [z] flags: current, []

Additional information

Used simplifier rules: in-15: ⊦ x + a ' = y → a ∈ y;%a: ⊦ (x + y) + z = x + y + z;%app-06: ⊦ x + a ' = y + b ' ↔ x = y ∧ a = b; Used terms: z