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Proof step 35 from proof array-exists-first-neq (conclusion)

⊦ ∃ n. n ≤ # ar0 ∧ n ≤ # ar1 ∧ subarray(ar0, 0, n) = subarray(ar1, 0, n) ∧ (n ≠ # ar0 ∧ n ≠ # ar1 → ar0[n] ≠ ar1[n])

User applied rule induction

And obtained the following new goals

Proof step 34:
true, ∀ ar, ar1. # ar < # ar0 → (true → (∃ n. n ≤ # ar ∧ n ≤ # ar1 ∧ subarray(ar, 0, n) = subarray(ar1, 0, n) ∧ (n ≠ # ar ∧ n ≠ # ar1 → ar[n] ≠ ar1[n]))) ⊦ ∃ n. n ≤ # ar0 ∧ n ≤ # ar1 ∧ subarray(ar0, 0, n) = subarray(ar1, 0, n) ∧ (n ≠ # ar0 ∧ n ≠ # ar1 → ar0[n] ≠ ar1[n])

Rule argument

[INDHYPARG : [STANDARDINDUCTIONTYPE : ], true, false, # ar0, [] → [], <]

Additional information

[[CONCRETESIMPRULES : ([CSIMPLESSPRED : <])], [EXTRATERMS : (# ar0)]]