[Imprint][Data Privacy Statement][E-Mail]



View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 4 from proof rax-02 (previous step)

i + 1 < 0, i < 0 ⊦ i→n(i + 1) = i→n(~ i) -1

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: Int2nat-pos seq: ⊦ 0 < i → i→n(i) = i→n(i - 1) + 1 subst: [i] → [~ i] flags: <all paths>

Additional information

Used simplifier rules: abs-neg: ⊦ i→n(~ i) = i→n(i);%abs-neg: ⊦ i→n(~ i) = i→n(i);% ⊦ ~ i - j = ~(i + j); (s)% ⊦ (m + 1)-1 = m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ n -1 = n - 1; (s)% ⊦ i + (* j) < (* j0) ↔ i < j0 - j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ 0 < ~ i ↔ i < 0; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < j → (j ≤ i ↔ false); (as) Used terms: ~ i