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Proof step 8 from proof extension (previous step)

# y = # y0, ∀ n0. ¬ # y0 < n0 → (a0 ' + y)[n0] = (a ' + y0)[n0], (n < # y0 → y[n] = y0[n]) → y = y0 ⊦ a0 ' + y = a ' + y0

User applied rule all left

And obtained the following new goals

Proof step 9:
# y = # y0, ¬ # y0 < 0 → (a0 ' + y)[0] = (a ' + y0)[0], ∀ n0. ¬ # y0 < n0 → (a0 ' + y)[n0] = (a ' + y0)[n0], (n < # y0 → y[n] = y0[n]) → y = y0 ⊦ a0 ' + y = a ' + y0

Rule argument

pos: left 2 terms: 0 abort: F discard: F computed: T

Additional information

Used terms: 0