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Proof step 4 from proof extension (previous step)

# x = 1 + # y0, ∀ x. # x = # y0 ∧ (∀ n. n < # y0 → x[n] = y0[n]) → x = y0, ∀ n. ¬ # y0 < n → x[n] = (a ' + y0)[n] ⊦ x = a ' + y0

User applied rule all left

And obtained the following new goals

Proof step 5:
# x = 1 + # y0, # x.tail = # y0 ∧ (∀ n. n < # y0 → x.tail[n] = y0[n]) → x.tail = y0, ∀ n. ¬ # y0 < n → x[n] = (a ' + y0)[n] ⊦ x = a ' + y0

Rule argument

pos: left 2 terms: x.tail abort: F discard: T computed: F

Additional information

Used terms: x.tail