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Proof step 1 from proof abs-succ (conclusion) (back to tree)

⊦ i < 0 → (i + 1 = abs(j) ↔ i = ~1 ∧ j = 0)

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof abs-succ (previous step) (back to tree)

¬(i + 1 = abs(j) ↔ i = ~1 ∧ j = 0), i < 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; add-c: ⊦ i + j = j + i; from specification int-basic2 add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2

Proof step 3 from proof abs-succ (previous step) (back to tree)

i + 1 = abs(j), i < 0 ⊦ i = ~1 ∧ j = 0

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Rule argument

fma: j < 0

Additional information

Used simplifier rules: Abs-pos: ⊦ ¬ j < 0 → abs(j) = j; succ-02: ⊦ i < j + 1 ↔ i ≤ j; le-def: ⊦ i ≤ j ↔ ¬ j < i; numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; mone-06: ⊦ ~1 < i ↔ 0 ≤ i; lels-03: ⊦ i < j → (j ≤ i ↔ false); lf-07: ⊦ ¬ i < i0 ∧ j < i0 → j < i; (forward) from specification int-basic1 ⊦ ¬ 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 ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss) Used formulas: j < 0

Proof step 4 from proof abs-succ (previous step) (back to tree)

i + 1 = abs(j), i < 0, j < 0 ⊦ i = ~1 ∧ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: abs-08: ⊦ i < 0 → abs(i) = ~ i; Minus-def: ⊦ ~ i = 0 - i; sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; add-c: ⊦ i + j = j + i; from specification int-basic2 add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2

Proof step 5 from proof abs-succ (previous step) (back to tree)

~1 = i + j, j ≠ 0, i < 0, j < 0 ⊦

User applied rule cut formula

And closed the goal!

Rule argument

fma: i = ~ j - 1

Additional information

Used simplifier rules: sub-add-01: ⊦ i + (j - i0) = i + j - i0; mone-01: ⊦ i + ~1 = i - 1; sub-add-09: ⊦ i - j + i = ~ j; Minus-def: ⊦ ~ i = 0 - i; sub-sub: ⊦ i - j - i0 = i - j + i0; num-subadd: ⊦ (* i) - j + (* i0) = i - i0 - j; sub-ls: ⊦ i - j < i0 ↔ i < j + i0; mone-06: ⊦ ~1 < i ↔ 0 ≤ i; lels-03: ⊦ i < j → (j ≤ i ↔ false); Minus-def: ⊦ ~ i = 0 - i; sub-sub: ⊦ i - j - i0 = i - j + i0; num-subadd: ⊦ (* i) - j + (* i0) = i - i0 - j; sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 add-c: ⊦ i + j = j + i; from specification int-basic2 ⊦ 0 + i = i; (s) Used formulas: i = ~ j - 1

Proof Tree for abs-succ (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