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Proof step 6 from proof monle-leq (previous step)

nσ[(m1 + 1) + m] < nσ[m], mon≤(nσ), ⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # nσ, ∀ n, m, nσ. m1 = n - m ∧ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ ∧ nσ[n] < nσ[m] → nσ[m] = 0 ⊦ nσ[m] = 0

User applied rule all left

And obtained the following new goals

Proof step 7:
nσ[(m1 + 1) + m] < nσ[m], mon≤(nσ), ⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # nσ, m1 = m1 + m - m ∧ mon≤(nσ) ∧ m ≤ m1 + m ∧ ⌜m1 + m⌝ < # nσ ∧ nσ[m1 + m] < nσ[m] → nσ[m] = 0 ⊦ nσ[m] = 0

Rule argument

pos: left 4 terms: m1 + m, m, nσ abort: F discard: T computed: F

Additional information

Used terms: m1 + m, m, nσ