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Proof step 3 from proof inapp-14 (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
⊦ x ≠ [] → (∃ y, a. x = y + a ')

Rule argument

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

Additional information

Used simplifier rules: app-06: ⊦ x + a ' = y + b ' ↔ x = y ∧ a = b;%neq: ⊦ ¬ a ∈ x → x ≠ y + a ' + z;%a: ⊦ (x + y) + z = x + y + z;%app: ⊦ x + [] = x; Used terms: z