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Proof step 10 from proof get-last-04 (previous step)

# x = m + 1, m < # x ⊦ x[m] = x.last, x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# x = m + 1 ⊦ x[m] = x.last, x = []

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s)% ⊦ m + 1 ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat