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Proof step 27 from proof from-append-add (previous step)

∀ n2. nf0(n2) = ((σ to n) + ⌜nf⌝ )[n2] ⊦ nf0(n0 + n1 + n) = nf(n0 + n1), # σ < ⌜n⌝

User applied rule all left

And obtained the following new goals

Proof step 28:
nf0(n0 + n1 + n) = ((σ to n) + ⌜nf⌝ )[n0 + n1 + n] ⊦ nf0(n0 + n1 + n) = nf(n0 + n1), # σ < ⌜n⌝

Rule argument

pos: left 1 terms: n0 + n1 + n abort: F discard: T computed: T

Additional information

Used terms: n0 + n1 + n