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Proof step 14 from proof firstn-last-01 (previous step)

a0 ' + x0 = firstn(n1, a0 ' + x0) + a ' + z, # x0 = n1 + # z, # z ≤ # x0, a0 ∈ firstn(n1, a0 ' + x0), x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → firstn(# x0 - # z, x0).last = a ⊦ n1 = 0, firstn(n1, x0).last = a

User applied rule case distinction

And obtained the following new goals

Proof step 15:
a0 ' + x0 = firstn(n1, a0 ' + x0) + a ' + z, # x0 = n1 + # z, # z ≤ # x0, a0 ∈ firstn(n1, a0 ' + x0) ⊦ x0 = firstn(# x0 - 1 + # z, x0) + a ' + z, n1 = 0, firstn(n1, x0).last = a
Proof step 16:
firstn(# x0 - # z, x0).last = a, a0 ' + x0 = firstn(n1, a0 ' + x0) + a ' + z, # x0 = n1 + # z, # z ≤ # x0, a0 ∈ firstn(n1, a0 ' + x0) ⊦ n1 = 0, firstn(n1, x0).last = a

Rule argument

pos: left 5

Proof step 14 from proof firstn-last-01 (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information