[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 11 from proof ex-ls-add (previous step)

i0 < j, ∀ j. i0 < j → (∃ i1. j = i0 + i1 ∧ 0 < i1) ⊦ i0 = j, ∃ i. j + 1 = i0 + i ∧ 0 < i

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: j abort: F discard: F computed: T

Additional information

Used simplifier rules: eq: ⊦ i = i + j ↔ j = 0;%add-a: ⊦ i + j + i0 = (i + j) + i0;%add-c: ⊦ i + j = j + i;%add-less: ⊦ 0 < i → i1 < i1 + i;%add-in: ⊦ i + j = i + i0 ↔ j = i0;%zero-02: ⊦ 0 < i ∧ 0 < j → 0 < i + j;%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1 Used terms: j