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Proof step 51 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 ⊦ ar[n] = a, m = n

User applied rule all left

And obtained the following new goals

Proof step 50:
ar[m] = ar[n], n < # ar, m < # ar, m < # ar → (∀ n. m < n ∧ n < # ar → ar[m] ≠ ar[n] ∨ ar[m] = a), ∀ 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 ⊦ ar[n] = a, m = n

Rule argument

[EXRARG : pos: left 4, [extquanttermlist: (m)abort:F, discard:F, computed:T]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (m)]]