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Proof step 9 from proof valid-idx-remlast (previous step)

ar[n, ar[n].idx:= 0][m1] = elemindex(a, m2), kar[m] = key-index(key, n), kar[0, key-index(key, n)][m0] = key-index(key0, m1), m0 + 1 < # kar, m < # kar, m0 < m, m0 < # kar, ¬ m < m0 → kar[m0].idx < # ar ∧ ar[kar[m0].idx].idx = m0, ∀ m0. ¬ m < m0 → kar[m0].idx < # ar ∧ ar[kar[m0].idx].idx = m0 ⊦ m1 < # ar ∧ m2 = m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
ar[kar[m0].idx].idx = m0, ar[n, ar[n].idx:= 0][m1] = elemindex(a, m2), kar[m] = key-index(key, n), kar[0, key-index(key, n)][m0] = key-index(key0, m1), kar[m0].idx < # ar, m0 + 1 < # kar, m < # kar, m0 < m, m0 < # kar, ∀ m0. ¬ m < m0 → kar[m0].idx < # ar ∧ ar[kar[m0].idx].idx = m0 ⊦ m1 < # ar ∧ m2 = m0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → ¬ n < m; (s)