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Proof step 2 from proof stutf-alltakenmax (previous step)

stutf(nσ), ⌜m⌝ < # nσ, n < nσ[m] ⊦ nσ[m] = 0, ∃ m0. m0 < m ∧ nσ[m0] = n ∧ nσ[m0 + 1] = nσ[m0] + 1

User applied rule induction

And obtained the following new goals

Proof step 3:
stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n < nσ[m], ∀ m0, nσ, n. m0 < m → stutf(nσ) ∧ ⌜m0⌝ < # nσ ∧ n < nσ[m0] → nσ[m0] = 0 ∨ (∃ m1. m1 < m0 ∧ nσ[m1] = n ∧ nσ[m1 + 1] = nσ[m1] + 1) ⊦ nσ[m] = 0 ∨ (∃ m0. m0 < m ∧ nσ[m0] = n ∧ nσ[m0 + 1] = nσ[m0] + 1)

Rule argument

indvar: m wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: m