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

ar[n0, ar[n0].idx:= m0][n1, ar[n0, ar0[n0].idx:= m0][n1].idx:= m1][m2] = elemindex(a, m3), kar[m0] = key-index(key1, n1), kar[m1] = key-index(key0, n0), kar[m0, key-index(key0, n0)][m1, key-index(key1, n1)][m] = key-index(key, m2), m < # kar, m < n, m0 < # kar, m1 < # kar, m0 < n, m1 < n, n ≤ # kar, m < n → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m, ∀ m. m < n → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m ⊦ m2 < # ar ∧ m3 = m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
ar[kar[m].idx].idx = m, ar[n0, ar[n0].idx:= m0][n1, ar[n0, ar0[n0].idx:= m0][n1].idx:= m1][m2] = elemindex(a, m3), kar[m0] = key-index(key1, n1), kar[m1] = key-index(key0, n0), kar[m0, key-index(key0, n0)][m1, key-index(key1, n1)][m] = key-index(key, m2), kar[m].idx < # ar, m < # kar, m < n, m0 < # kar, m1 < # kar, m0 < n, m1 < n, n ≤ # kar, ∀ m. m < n → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m ⊦ m2 < # ar ∧ m3 = m

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