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Proof step 48 from proof challenge3b (previous step)

ar[m] = ar[n], n < # ar, m < # ar, ∀ m1. m1 < # ar → (∀ 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 ∧ n < # ar → ar[m] ≠ ar[n] ∨ ar[m] = a ⊦ ar[n] = a, m = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 47:
ar[m] = ar[n], n < # ar, n < m, m < # ar, n + 1 < # ar, ar[m1] ≠ ar[n2] ∨ ar[m1] = a ∨ m1 = n2 ∨ ¬ m1 < # ar ∨ ¬ n2 < # ar, ∀ m1. m1 < # ar → (∀ n. m1 < n ∧ n < # ar → ar[m1] ≠ ar[n] ∨ ar[m1] = a) ⊦ ar[n] = a, m = n

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m +1 = m + 1], [CSIMPSEQ : ⊦ m < n → n ≠ 0], [CSIMPFORWARD : ⊦ n < n0 ∧ n0 < n1 → n < n1], [CSIMPFORWARD : ⊦ m < n → n ≠ 0], [CSIMPFORWARD : ⊦ m < n ∧ n < n0 → m +1 < n0], [CSIMPSEQ : ⊦ m ≠ n → (¬ m < n ↔ n < m)])]]