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Proof step 4 from proof ex-ls-add (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: j -1 abort: F discard: F computed: F

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1;%add-a: ⊦ i + j + i0 = (i + j) + i0;%add-c: ⊦ i + j = j + i;%lst-03: ⊦ i + j < i0 ∧ ¬ j < j0 → i + j0 < i0;%ls-01: ⊦ i + j < i + i0 ↔ j < i0;%add-less: ⊦ 0 < i → i1 < i1 + i;%add-in: ⊦ i + j = i + i0 ↔ j = i0;%ls-pred: ⊦ i < j -1 ↔ i +1 < j; from specification int-basic1%pred: ⊦ i = j -1 ↔ i +1 = j; from specification int-basic1%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1 Used terms: j -1