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Proof step 2 from proof prefix-intro (previous step)

x + x0 = y + y0, # x ≤ # y ⊦ x ⊑ y

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 3:
n ≤ # x ⊦ ∃ y, z. x = y + z ∧ # y = n

Rule argument

spec: list inst: name: split-first seq: n ≤ # x ⊦ ∃ y, z. x = y + z ∧ # y = n subst: [n, x] → [# x, y] flags:

Additional information

Used simplifier rules: prefix-app-05: ⊦ x ⊑ x + y;%len-01: ⊦ # x = # z → (x + y = z + y0 ↔ x = z ∧ y = y0); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list%lep: ⊦ m ≤ m + n; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: # x, y