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Proof step 2 from proof in-09 (previous step)

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

User applied rule insert spec-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:

Additional information

Used simplifier rules: f-01: ⊦ dups(x + a ' + y) ↔ a ∈ x ∨ a ∈ y ∨ dups(x + y);%app: ⊦ x + [] = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%app-06: ⊦ x + a ' = y + b ' ↔ x = y ∧ a = b; from specification list%in-empty: ⊦ ¬ a ∈ []; from specification list%ni: ⊦ x + a ' = y + z ∧ ¬ a ∈ z → z = []; (forward) from specification list%in-15: ⊦ x + a ' = y → a ∈ y; (forward) from specification list%e-02: ⊦ x + a ' ≠ []; from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%in-17: ⊦ a ∈ b ' ↔ a = b; from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list Used terms: z