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Proof step 4 from proof nat (previous step)

i < 0, j < 0 ⊦ i→n(i) / i→n(j) = i→n(i / j), j = 0

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: natpos seq: 0 < j, 0 ≤ i ⊦ i→n(i) / i→n(j) = i→n(i / j) subst: [j, i] → [~ j, ~ i]

Additional information

Used simplifier rules: abs-neg: ⊦ i→n(~ i) = i→n(i);% ⊦ ~ i = 0 ↔ i = 0; (s)%j ≠ 0 ⊦ ~ i / j = ~(i / j); (s)%j ≠ 0 ⊦ i / ~ j = ~(i / j); (s)% ⊦ ~ ~ i = i; (s)% ⊦ 0 < ~ i ↔ i < 0; (s)% ⊦ i ≤ ~ j ↔ i + j ≤ 0; (s)% ⊦ i + 0 = i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < j → ¬ j < i; (s) Used terms: ~ j, ~ i