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

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

User applied rule structural induction

And obtained the following new goals

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

Rule argument

indvar: m

Additional information

Used simplifier rules: stutf-zero: ⊦ stutf(nσ) → nσ[0] = 0;%nat freely generated by 0, +1;;%zero-len: ⊦ ⌜0⌝ < # nσ ↔ nσ ≠ []; from specification stream[natstream]%lez-01: ⊦ m ≤ 0 ↔ m = 0; from specification nat