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Proof step 6 from proof lastN-ex (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 7:
# x = n0 + n, n ≠ 0 → x = firstn(n0, x) + restn(n0, x) ⊦ n = 0, ∃ y. x = y + lastn(n, x) ∧ # y = n0

Rule argument

fma: n = 0

Additional information

Used simplifier rules: lastN-zero: ⊦ lastn(0, x) = [];% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ x + [] = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list Used formulas: n = 0