[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 2 from proof abs-succ-01 (previous step)

⊦ i→n(i + 1) < i→n(i) ↔ i < 0

User applied rule case distinction

And obtained the following new goals

Proof step 3:
i < 0 ⊦ i→n(i + 1) < i→n(i)

Rule argument

pos: right 1

Additional information

Used simplifier rules: rax-04: ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1;% ⊦ i < j ↔ ¬ j ≤ i; (ss)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic