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Proof step 14 from proof challenge3a (previous step)

m2 = m0, # ar = n, ¬ bv0, ar1 = ar, ar[m] = ar[n0], m ≠ n0, m0 < # ar, m < # ar, n0 < # ar, ∀ n0. m0 < n0 ∧ n0 < # ar → ar[n0] ≠ ar[m0], ∀ m1. m1 < m0 → (∀ n. m1 < n ∧ n < # ar → ar[m1] ≠ ar[n]) ⊦ ⦉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)

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 13:
m2 = m0, # ar = n, ¬ bv0, ar1 = ar, ar[m] = ar[n0], m ≠ n0, m0 < # ar, m < # ar, n0 < # ar, ∀ n0. m0 < n0 ∧ n0 < # ar → ar[n0] ≠ ar[m0], ∀ m1. m1 < m0 → (∀ n. m1 < n ∧ n < # ar → ar[m1] ≠ ar[n]), (bv0 ⊃ (ar[m1] ≠ ar[n2] ∨ m1 = n2 ∨ ¬ m1 < # ar ∨ ¬ n2 < # ar); (∃ m1. ¬ (m1 < m2 + 1 → (∀ n. m1 < n ∧ n < # ar → ar[m1] ≠ ar[n])))) ∨ ¬ m2 + 1 ≤ # ar ∨ (∀ m, n. ¬ (ar1[m] = ar1[n] ∧ m ≠ n ∧ m < # ar1 ∧ n < # ar1)) ∨ m2 + 1 ≤ m0 ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]