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Proof step 7 from proof challenge3a (previous step)

ar[m] = ar[n0], ar[m1] = ar[n], m1 ≤ m0, n0 < # ar, m < # ar, m0 < # ar, n < # ar, m1 < n, m1 < # ar, m1 + 1 < # ar, m1 < m0 → (∀ n. m1 < n ∧ n < # ar → ar[m1] ≠ ar[n]), ∀ m1. m1 < m0 → (∀ n. m1 < n ∧ n < # ar → ar[m1] ≠ ar[n]), ∀ n0. m0 < n0 ∧ n0 < # ar → ar[n0] ≠ ar[m0] ⊦ m1 = m0, m = n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
ar[m1] = ar[n], ar[m] = ar[n0], m1 < m0, m1 + 1 < # ar, m1 < # ar, m1 < n, n < # ar, m0 < # ar, m < # ar, n0 < # ar, ∀ n. m1 < n ∧ n < # ar → ar[m1] ≠ ar[n], ∀ n0. m0 < n0 ∧ n0 < # ar → ar[n0] ≠ ar[m0], ∀ m1. m1 < m0 → (∀ n. m1 < n ∧ n < # ar → ar[m1] ≠ ar[n]) ⊦ m = n0, m1 = m0

Rule argument

[emptyarg]

Additional information

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