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Proof step 3 from proof stutf-notbig (previous step)

n0 + 1 < nσ[n0 + 1], stutf(nσ), ⌜n0⌝ + ⌜1⌝ < # nσ, ∀ nσ0. stutf(nσ0) ∧ ⌜n0⌝ < # nσ0 → ¬ n0 < nσ0[n0] ⊦ nσ[n0 + 1] = 0

User applied rule all left

And obtained the following new goals

Proof step 4:
n0 + 1 < nσ[n0 + 1], stutf(nσ), ⌜n0⌝ + ⌜1⌝ < # nσ, stutf(nσ) ∧ ⌜n0⌝ < # nσ → ¬ n0 < nσ[n0] ⊦ nσ[n0 + 1] = 0

Rule argument

pos: left 4 terms: nσ abort: F discard: T computed: T

Additional information

Used terms: nσ