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Proof step 3 from proof rev-at (previous step)

# ar = n0 + n + 1 ∧ n + 1 ≤ # ar, n < # ar ⊦ rev(ar)[n] = ar[n0], # ar = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# ar = n0 + n + 1, n < # ar ⊦ # ar = 0, rev(ar)[n] = ar[n0]

Rule argument

Additional information

Used simplifier rules: l-02: ⊦ ¬ m + n < m; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (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%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat