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Proof step 28 from proof challenge3a (previous step)

m0 < n1 ∧ n1 ≤ # ar ∧ (m0 < n3 ∧ n3 < n1 → ar[n3] ≠ ar[m0]) → ⦉m3 := m0, bv := false, ar0 := ar ; n4 := n2, m7 := m1 ; while n1 < # ar0 ∧ ¬ bv do if ar0[m3] = ar0[n1] then m7 := m3, n4 := n1, bv := true else n1 := n1 + 1⦊ ⦉m3 := m3 + 1⦊ ( (bv ⊃ (ar[m7] = ar[n4] ∧ m7 ≠ n4 ∧ m7 < # ar ∧ n4 < # ar); (∀ m1. m1 < m3 → (∀ n. m1 < n ∧ n < # ar → ar[m1] ≠ ar[n]))) ∧ m3 ≤ # ar ∧ (∃ m, n. ar0[m] = ar0[n] ∧ m ≠ n ∧ m < # ar0 ∧ n < # ar0) ∧ ¬ m3 ≤ m0), m2 = m0, n1 = n + 1, ar[m] = ar[n0], # ar = (n5 + 1) + n, ar1 = ar, ¬ bv0, ar[m0] ≠ ar[n], m ≠ n0, m0 < n5 + n, m ≤ n5 + n, n0 ≤ n5 + n, m0 < # ar, m0 < n, m < # ar, n0 < # ar, n < # ar, ∀ n0. m0 < n0 ∧ n0 < n → ar[n0] ≠ ar[m0], ∀ m1. m1 < m0 → (∀ n. m1 < n ∧ n < # ar → ar[m1] ≠ ar[n]) ⊦ ⦉while n1 < # ar1 ∧ ¬ bv0 do if ar1[m2] = ar1[n1] then m1 := m2, n2 := n1, bv0 := true else n1 := n1 + 1⦊ ⦉m2 := m2 + 1⦊ ( (bv0 ⊃ (ar[m1] = ar[n2] ∧ m1 ≠ n2 ∧ m1 < # ar ∧ n2 < # ar); (∀ m1. m1 < m2 → (∀ n. m1 < n ∧ n < # ar → ar[m1] ≠ ar[n]))) ∧ m2 ≤ # ar ∧ (∃ m, n. ar1[m] = ar1[n] ∧ m ≠ n ∧ m < # ar1 ∧ n < # ar1) ∧ ¬ m2 ≤ m0)

User applied rule case distinction

And obtained the following new goals

Rule argument

[CASEDARG : pos: left 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]