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Proof step 10 from proof list-split (previous step)

# y2 ≤ # y0, y2 ⊑ y0 + z, ∀ x, z. # x ≤ # y0 ∧ x ⊑ y0 + z → (∃ y1. y0 = x + y1) ⊦ ∃ y. y0 = y2 + y

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: y2, z

Additional information

Used simplifier rules: ⊦ #(x + y) = # x + # y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ m ≤ m + n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ x ⊑ x + y; (s)% ⊦ x + y = x + z ↔ y = z; (s) Used terms: y2, z