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Proof step 2 from proof inapp-11 (previous step)

x + y = z + a ', b ∈ x, a ∈ x ∨ a ∈ y ⊦ a = b, b ∈ 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] → [y] flags: current, []

Additional information

Used simplifier rules: inapp-12: ⊦ x + y = z + a ' ∧ b ∈ x ∧ y ≠ [] → b ∈ z;%a: ⊦ (x + y) + z = x + y + z;%app: ⊦ x + [] = x;%in-empty: ⊦ ¬ a ∈ [];%in-15: ⊦ x + a ' = y → a ∈ y;%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y);%in-17: ⊦ a ∈ b ' ↔ a = b; Used terms: y