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Proof step 5 from proof id-idx-trans (previous step)

# ar2 = # ar0, # ar1 = # ar0, ar0[n] = elemindex(a, m), ar2[n] = elemindex(a0, m0), n < # ar0, ∀ n. n < # ar0 → ar1[n].elem = ar2[n].elem, ∀ n. n < # ar0 → ar0[n].elem = ar1[n].elem ⊦ a = a0

User applied rule all left

And obtained the following new goals

Proof step 6:
# ar1 = # ar0, # ar2 = # ar0, ar2[n] = elemindex(a0, m0), ar0[n] = elemindex(a, m), n < # ar0, n < # ar0 → ar0[n].elem = ar1[n].elem, ∀ n. n < # ar0 → ar1[n].elem = ar2[n].elem ⊦ a = a0

Rule argument

pos: left 7 terms: n abort: F discard: T computed: T

Additional information

Used terms: n