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Proof step 4 from proof stutf-alltaken (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 5:
stutf(nσ), ⌜m0⌝ + ⌜1⌝ < # nσ, n ≤ nσ[m0 + 1], stutf(nσ) ∧ ⌜m0⌝ < # nσ ∧ n ≤ nσ[m0] → (∃ m1. m1 ≤ m0 ∧ nσ[m1] = n), ∀ nσ, n. stutf(nσ) ∧ ⌜m0⌝ < # nσ ∧ n ≤ nσ[m0] → (∃ m1. m1 ≤ m0 ∧ nσ[m1] = n) ⊦ ∃ m. m ≤ m0 + 1 ∧ nσ[m] = n

Rule argument

pos: left 4 terms: nσ, n abort: F discard: F computed: F

Additional information

Used terms: nσ, n