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Proof step 7 from proof split-first (previous step)

# y = n - 1, n ≤ ((n - 1) + # z) + 1 ⊦ ∃ y0, z0. a ' + y + z = y0 + z0 ∧ # y0 = n

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ n ≤ n0 → (m + (n0 - n)) + n = m + n0
Proof step 9:
1 ≤ n, # y = n - 1, n ≤ # z + n ⊦ ∃ y0, z0. a ' + y + z = y0 + z0 ∧ # y0 = n

Rule argument

spec: nat inst: name: sub-add seq: ⊦ n ≤ n0 → (m + (n0 - n)) + n = m + n0 subst: [m, n, n0] → [# z, 1, n] flags: rewrite

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z;%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%lez: ⊦ 0 ≤ m; from specification nat%l: ⊦ # x = 0 ↔ x = []; from specification list-data%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: # z, 1, n