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Proof step 54 from proof challenge3b (previous step)

¬ (m2 < # ar1 ∧ ¬ bv0), 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, ar[m1] ≠ ar[n2] ∨ ar[m1] = a ∨ m1 = n2 ∨ ¬ m1 < # ar ∨ ¬ n2 < # ar ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 53:
ar[m] = ar[n], m2 ≤ # ar, m < # ar, n < # ar, bv0 ⊃ (ar[m1] = ar[n2] ∧ ar[m1] ≠ a ∧ m1 ≠ n2 ∧ m1 < # ar ∧ n2 < # ar) ; (∀ m1. m1 < m2 → (∀ n. m1 < n ∧ n < # ar → ar[m1] ≠ ar[n] ∨ ar[m1] = a)), ar[m1] ≠ ar[n2] ∨ ar[m1] = a ∨ m1 = n2 ∨ ¬ m1 < # ar ∨ ¬ n2 < # ar ⊦ m = n, ar[n] = a, m2 < # ar ∧ ¬ bv0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]