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Proof step 2 from proof neq-03 (previous step)

y + b ' = y0 + c ' + z ⊦ c = b, c ∈ y

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: app: ⊦ x + [] = x;%a: ⊦ (x + y) + z = x + y + z;%app-06: ⊦ x + a ' = y + b ' ↔ x = y ∧ a = b;%in-15: ⊦ x + a ' = y → a ∈ y;%neq: ⊦ ¬ a ∈ x → x ≠ y + a ' + z; Used terms: z