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Proof step 3 from proof pls (previous step)

i0 < j0, i1 + 1 < j, ∀ j, i0, j0. i1 < j ∧ i0 < j0 → i1 + i0 < j + j0 ⊦ (i1 + 1) + i0 < j + j0

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: j -1, i0, j0 abort: F discard: T 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;%add-pred: ⊦ i + j -1 = (i + j)-1;%ls-pred: ⊦ i < j -1 ↔ i +1 < j; from specification int-basic1 Used terms: j -1, i0, j0