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Proof step 2 from proof inex (previous step)

x + a ' = y + b ' + z ⊦ a = b, ∃ z0. z = z0 + a '

User applied rule insert lemma

And obtained the following new goals

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;%e-02: ⊦ x + a ' ≠ [];%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y);%in-17: ⊦ a ∈ b ' ↔ a = b; Used terms: z