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Proof step 9 from proof list-split (previous step)

x = a + y2, # x ≤ # y0 + 1, y2 ⊑ y0 + z, ∀ x, z. # x ≤ # y0 ∧ x ⊑ y0 + z → (∃ y1. y0 = x + y1) ⊦ x = [], ∃ y. a ' + y0 = x + y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# y2 ≤ # y0, y2 ⊑ y0 + z, ∀ x, z. # x ≤ # y0 ∧ x ⊑ y0 + z → (∃ y1. y0 = x + y1) ⊦ ∃ y. y0 = y2 + y

Rule argument

Additional information

Used simplifier rules: ⊦ a ' + x = a ' + y ↔ x = y; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m +1 = m + 1; (s)% ⊦ #(a ' + x) = (# x)+1; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)