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Proof step 5 from proof inapp-13 (previous step)

a ' + x = y + b ', a ∈ y, b ∈ x, c ∈ x ⊦ c = b, y = [], c ∈ y

User applied rule insert lemma

And obtained the following new goals

Proof step 6:
⊦ x ≠ [] → (∃ y, a. x = y + a ')

Rule argument

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

Additional information

Used simplifier rules: app: ⊦ x + [] = x;%a: ⊦ (x + y) + z = x + y + z;%c-05: ⊦ a ' = x + b ' ↔ a = b ∧ x = [];%in-15: ⊦ x + a ' = y → a ∈ y;%app-06: ⊦ x + a ' = y + b ' ↔ x = y ∧ a = b;%e-02: ⊦ x + a ' ≠ [];%in-append: ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y;%in-17: ⊦ a ∈ b ' ↔ a = b;%in-14: ⊦ a ' + x = y → a ∈ y;%e-01: ⊦ a ' + x ≠ []; Used terms: x