[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 9 from proof int-simp-04 (previous step)

i0 + i < 0, 0 < i, 0 < i0 ⊦

User applied rule structural induction

And obtained the following new goals

Proof step 10:
0 < i0, 0 < i1 +1, i0 + i1 +1 < 0, ∀ i0. ¬(i0 + i1 < 0 ∧ 0 < i1 ∧ 0 < i0) ⊦

Rule argument

indvar: i

Additional information

Used simplifier rules: sub-pred: ⊦ i -1 = i - 1;%sub-add-01: ⊦ i + (j - i0) = i + j - i0;%sub-ls: ⊦ i - j < i0 ↔ i < j + i0;%lsone: ⊦ i < 1 ↔ i ≤ 0;%le-def: ⊦ i ≤ j ↔ ¬ j < i;%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j;%numlsf-01: ⊦ (* i) < j ∧ 0 < i → 0 ≤ j ∧ 0 < j;%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j;%int generated by 0, +1, -1;;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1% ⊦ i < j → ¬ j < i; (s)% ⊦ 0 < i ∧ 0 < j → 0 < i + j; (s)% ⊦ 0 + i = i; (s)