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Proof step 51 from proof subarray-eq-elim (previous step)

# ar = (n1 + m) + n ⊦ ∃ ar0, ar1. ar = ar0 + subarray(ar, # ar0, m) + ar1 ∧ # ar0 = n

User applied rule exists right

And obtained the following new goals

Proof step 52:
# ar = (n1 + m) + n ⊦ ar = subarray(ar, 0, n) + subarray(ar, # subarray(ar, 0, n), m) + subarray(ar, n + m, n1) ∧ # subarray(ar, 0, n) = n

Rule argument

pos: right 1 terms: subarray(ar, 0, n), subarray(ar, n + m, n1) abort: F discard: T computed: F

Additional information

Used terms: subarray(ar, 0, n), subarray(ar, n + m, n1)