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Proof step 23 from proof valid-idx-xhg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
ar0[n0].idx:= m0.idx:= m0 = elemindex(a, m3), kar[m0] = key-index(key1, n0), ar[n0] = elemindex(a1, m0), m0 < # kar, m0 < n, n0 < # ar, n ≤ # kar, ∀ m. m < n → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m ⊦ m3 = m0

Rule argument

Additional information

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