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Proof step 19 from proof min-rec (previous step)

min(i ' + ints) ≤ i, min(ints) ≤ i, ∀ i0. i0 ∈ ints → min(i ' + ints) ≤ i0, ∀ i. i ∈ ints → min(ints) ≤ i ⊦ ints = [], min(i ' + ints) = min(ints)

User applied rule all left

And obtained the following new goals

Proof step 20:
min(i ' + ints) ≤ i, min(ints) ≤ i, min(ints) ∈ ints → min(i ' + ints) ≤ min(ints), ∀ i0. i0 ∈ ints → min(i ' + ints) ≤ i0, ∀ i. i ∈ ints → min(ints) ≤ i ⊦ ints = [], min(i ' + ints) = min(ints)

Rule argument

pos: left 3 terms: min(ints) abort: F discard: F computed: F

Additional information

Used terms: min(ints)