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Proof step 3 from proof in-09 (previous step)

x + a ' = y + b ' + z, b ∈ z, a ∈ y ∨ a ∈ z, b ∈ x ∨ b = a ⊦ dups(x), a ∈ x

User applied rule insert spec-lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: in-01-02: ⊦ a ∈ y → dups(x + a ' + y);% ⊦ x + [] = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + a ' = y + b ' ↔ x = y ∧ a = b; (ws)% ⊦ ¬ a ∈ []; (s)% ⊦ ¬ a ∈ x → (x + a ' = y + a ' + z ↔ y = x ∧ z = []); (ws)%ni: ⊦ x + a ' = y + z ∧ ¬ a ∈ z → z = []; (forward) from specification list%in-15: ⊦ x + a ' = y → a ∈ y; (forward) from specification list% ⊦ x + a ' ≠ []; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ x = y + a ' + z → a ∈ x; (s)%in-13: ⊦ x = y + a ' + z → a ∈ x; (forward) from specification list Used terms: z