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Proof step 52 from proof challenge3a (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 51:
ar[m] = ar[n], n < # ar, m < # ar, ∀ m1. m1 < # ar → (∀ n. m1 < n ∧ n < # ar → ar[m1] ≠ ar[n]), ar[m1] ≠ ar[n2] ∨ m1 = n2 ∨ ¬ m1 < # ar ∨ ¬ n2 < # ar ⊦ m = n

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m ≤ m], [CSIMPFORWARD : ⊦ m < n ∧ n = n0 → m < n0], [CSIMPSEQ : ⊦ m ≤ n → (n ≤ m ↔ m = n)], [CSIMPSEQ : ⊦ ¬ m < n ↔ n ≤ m])]]