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Proof step 28 from proof valid-idx-xhg (previous step)

ar[n0, ar0[n0].idx:= m0][n1].idx:= m1 = elemindex(a, m3), kar[m5] = key-index(key2, m4), ar[m4] = elemindex(a0, m5), ar[n1] = elemindex(a1, m0), ar[n0] = elemindex(a2, m1), kar[m0, key-index(key0, n0)][m5] = key-index(key, n1), kar[m1] = key-index(key0, n0), kar[m0] = key-index(key1, n1), m1 < # kar, m5 < n, m4 < # ar, n1 < # ar, m5 < # kar, m1 < n, n0 < # ar, m0 < # kar, m0 < n, n ≤ # kar, ∀ m. m < n → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m ⊦ m3 = m5, m5 = m1

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: m5 = m0

Additional information

Used simplifier rules: ⊦ m ≠ n → kar[n, ki][m] = kar[m]; (s)% ⊦ 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 < # kar → kar[m, ki][m] = ki; (s)% ⊦ m < # ar → ar[m, ei][m] = ei; (s) Used formulas: m5 = m0