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Proof step 12 from proof firstn-append-01 (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 13:
# x0 = n1 + # z, x0 = firstn(n1 - 1, x0) + a ' + z, # z ≤ # x0, ∀ z, a. x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → x0 = firstn(# x0 - # z, x0) + z ⊦ x0 = firstn(# x0 - 1 + # z, x0) + a ' + z, x0 = firstn(n1, x0) + z, n1 = 0
Proof step 18:
x0 = firstn(# x0 - # z, x0) + z, # x0 = n1 + # z, x0 = firstn(n1 - 1, x0) + a ' + z, # z ≤ # x0, ∀ z, a. x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → x0 = firstn(# x0 - # z, x0) + z ⊦ x0 = firstn(n1, x0) + z, n1 = 0

Rule argument

pos: left 4

Proof step 12 from proof firstn-append-01 (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information