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Proof step 4 from proof list-split (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# x ≤ # y0 + 1, x ⊑ (a ' + y0) + z, ∀ x, z. # x ≤ # y0 ∧ x ⊑ y0 + z → (∃ y1. y0 = x + y1) ⊦ ∃ y. a ' + y0 = x + y

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ #(a ' + x) = (# x)+1; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)