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Proof step 21 from proof valid-idx-xhg (previous step)

ar[n0, elemindex(a2, 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)][m5] = key-index(key, m2), ar[n0] = elemindex(a2, m1), ar[n1] = elemindex(a1, m0), ar[m4] = elemindex(a0, m5), kar[m5] = key-index(key2, m4), m5 = m1, m0 < n, m0 < # kar, n0 < # ar, m1 < n, m5 < # kar, n1 < # ar, m4 < # ar, m5 < n, m1 < # kar, n ≤ # kar, ∀ m. m < n → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m ⊦ m2 < # ar ∧ m3 = m5

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
ar[n0, ar0[n0].idx:= m0][m2].idx:= m1 = elemindex(a, m3), ar[m2] = elemindex(a1, m0), ar[n0] = elemindex(a0, m1), kar[m0] = key-index(key1, m2), kar[m1] = key-index(key0, n0), m1 < # kar, m1 < n, n0 < # ar, m0 < # kar, m0 < n, m2 < # ar, n ≤ # kar, ∀ m. m < n → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m ⊦ m3 = m1

Rule argument

Additional information

Used simplifier rules: ⊦ elemindex(a, m) = elemindex(a0, m0) ↔ a = a0 ∧ m = m0; (ws)%m < # ar ⊦ ar[m, ei][m] = ei; (s)% ⊦ # ar[n, ei] = # ar; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ key-index(key, m) = key-index(key0, m0) ↔ key = key0 ∧ m = m0; (ws)%m < # kar ⊦ kar[m, ki][m] = ki; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ # kar[n, ki] = # kar; (s)