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Proof step 14 from proof rev-at-h (previous step)

# ar = n2 + n1 + 1, n1 < # ar, ∀ ar, n0, m. # ar = n0 + m + 1 ∧ m < # ar ∧ m < n1 ∧ n1 ≤ # ar → rev(ar, n1)[m] = ar[n0] ⊦ rev(ar, n1)[n1, ar[n2]][n1] = ar[n2]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ m < # ar → ar[m, a][m] = a
Proof step 16:
# ar = n2 + n1 + 1, n1 < # ar, ∀ ar, n0, m. # ar = n0 + m + 1 ∧ m < # ar ∧ m < n1 ∧ n1 ≤ # ar → rev(ar, n1)[m] = ar[n0] ⊦ rev(ar, n1)[n1, ar[n2]][n1] = ar[n2], n1 < # rev(ar, n1)

Rule argument

spec: array name: Put-same seq: ⊦ m < # ar → ar[m, a][m] = a subst: [a, m, ar] → [ar[n2], n1, rev(ar, n1)] flags: <all paths>

Additional information

Used terms: ar[n2], n1, rev(ar, n1)