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Proof step 60 from proof challenge3a (conclusion)

∃ m, n. ar[m] = ar[n] ∧ m ≠ n ∧ m < # ar ∧ n < # ar ⊦ ⦉FINDDUP#(ar; m0, n0)⦊ (ar[m0] = ar[n0] ∧ m0 ≠ n0 ∧ m0 < # ar ∧ n0 < # ar)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 59:
ar[m] = ar[n], m ≠ n, n < # ar, m < # ar ⊦ ⦉FINDDUP#(ar; m0, n0)⦊ (ar[m0] = ar[n0] ∧ m0 ≠ n0 ∧ m0 < # ar ∧ n0 < # ar)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m < n → n ≠ 0], [CSIMPFORWARD : ⊦ m < n → n ≠ 0])]]