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Proof step 5 from proof lastN-ex (previous step)

# x - n < # x ↔ x = firstn(# x - n, x) + restn(# x - n, x) ∧ # x - n < # x, # x = n0 + n ⊦ ∃ y. x = y + lastn(n, x) ∧ # y = n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# x = n0 + n, n ≠ 0 → x = firstn(n0, x) + restn(n0, x) ⊦ ∃ y. x = y + lastn(n, x) ∧ # y = n0

Rule argument

Additional information

Used simplifier rules: eq-01: ⊦ m + n - m = n; from specification nat%l-01: ⊦ m < m + n ↔ n ≠ 0; from specification nat-basic2%ls-01: n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat