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Proof step 1 from proof ex-le-add (conclusion) (back to tree)

⊦ i ≤ j ↔ (∃ i0. j = i + i0 ∧ 0 ≤ i0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ i ≤ j ↔ ¬ j < i; add-c: ⊦ i + j = j + i; from specification int-basic2 add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 ⊦ i + j < j ↔ i < 0; (s)

Proof step 2 from proof ex-le-add (previous step) (back to tree)

¬ j < i ⊦ ∃ i0. j = i + i0 ∧ ¬ i0 < 0

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: j - i abort: F discard: T computed: F

Additional information

Used simplifier rules: sub-add-01: ⊦ i + (j - i0) = i + j - i0; sub-add: ⊦ i + j - i = j; sub-ls: ⊦ i - j < i0 ↔ i < j + i0; ⊦ 0 + i = i; (s) add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: j - i

Proof Tree for ex-le-add (View conclusion)

The following image shows the proof tree of sequent calculus. The proved goal is at the bottom. You can click on the nodes to view additional rules. See here for details on colors in a proof tree.

the proof tree