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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ref-1: ⊦ ¬ n < n; from specification nat-basic1%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%les: ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); from specification nat%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat