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Proof step 1 from proof append-size-less (conclusion)

⊦ # x < # x1 → (x + x0 = x1 + x2 ↔ (∃ y. x1 = x + y ∧ x0 = y + x2))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
x + x0 = x1 + x2, # x < # x1 ⊦ x1 = [], ∃ y. x1 = x + y ∧ x0 = y + x2

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z;%l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1