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Proof step 3 from proof challenge3b (previous step)

¬ bv0, ar[m] = ar[n], ar1 = ar, a1 = a, m2 = 0, m ≠ n, ar[n] ≠ a, n < # ar, m < # ar, (bv0 ⊃ (ar1[m1] ≠ ar1[n2] ∨ ar[m1] = a1 ∨ m1 = n2 ∨ ¬ m1 < # ar1 ∨ ¬ n2 < # ar1) ; (∃ m1. ¬ (m1 < m2 → (∀ n. m1 < n ∧ n < # ar1 → ar1[m1] ≠ ar1[n] ∨ ar[m1] = a1))) ) ∨ ¬ m2 ≤ # ar1 ∨ ar1 ≠ ar ∨ a1 ≠ a ∨ (∀ m, n. ¬ (ar1[m] = ar1[n] ∧ ar[m] ≠ a1 ∧ m ≠ n ∧ m < # ar1 ∧ n < # ar1)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
ar[m] = ar[n], n < # ar, m < # ar, ∀ m0, n0. ar[m0] ≠ ar[n0] ∨ ar[m0] = a ∨ m0 = n0 ∨ ¬ m0 < # ar ∨ ¬ n0 < # ar ⊦ ar[n] = a, m = n

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ 0 ≤ m], [CSIMPSEQ : ⊦ ¬ m < 0])]]