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Proof step 11 from proof arr2list-get (previous step)

m0 + 1 < # ar, n < # ar, n ≤ m0, ∀ n, ar. n < # ar ∧ n ≤ m0 ∧ m0 < # ar → (arr2list(ar, m0)[n] ↔ ar[n]) ⊦ n = m0 + 1, (arr2list(ar, m0) + ar[m0 + 1] ')[n] ↔ ar[n]

User applied rule all left

And obtained the following new goals

Proof step 10:
m0 + 1 < # ar, n < # ar, n ≤ m0, n < # ar ∧ n ≤ m0 ∧ m0 < # ar → (arr2list(ar, m0)[n] ↔ ar[n]), ∀ n, ar. n < # ar ∧ n ≤ m0 ∧ m0 < # ar → (arr2list(ar, m0)[n] ↔ ar[n]) ⊦ n = m0 + 1, (arr2list(ar, m0) + ar[m0 + 1] ')[n] ↔ ar[n]

Rule argument

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

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (n, ar)]]