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Proof step 3 from proof min-del-nlz-01 (previous step)

min((i ' + ints0) -l i0) < 0, min((i ' + ints0) -l i0) < min(i ' + ints0), 0 < min(i ' + ints0), 0 < min(ints0) → (∀ i0. ¬ min(ints0 -l i0) < 0) ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ints ≠ [] → min(i ' + ints) = min(i, min(ints))
Proof step 5:
¬ ints0 ≠ [], min((i ' + ints0) -l i0) < 0, min((i ' + ints0) -l i0) < min(i ' + ints0), 0 < min(i ' + ints0), 0 < min(ints0) → (∀ i0. ¬ min(ints0 -l i0) < 0) ⊦
Proof step 8:
ints0 ≠ [], min((i ' + ints0) -l i0) < 0, min((i ' + ints0) -l i0) < min(i, min(ints0)), 0 < min(i, min(ints0)), 0 < min(ints0) → (∀ i0. ¬ min(ints0 -l i0) < 0) ⊦

Rule argument

spec: inst: name: min-rec seq: ⊦ ints ≠ [] → min(i ' + ints) = min(i, min(ints)) subst: [ints, i] → [ints0, i] flags: rewrite, current

Additional information

Used terms: ints0, i