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Proof step 5 from proof max_exp-pos-03 (previous step)

max_exp(i0, j0) + max_exp(i, j) = 0 ⊦ max_exp(i0, j0) = 0 ∧ max_exp(i, j) = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ 0 ≤ i ∧ 0 ≤ j → (i + j = 0 ↔ i = 0 ∧ j = 0)

Rule argument

spec: int-abs inst: name: zero-eq seq: ⊦ 0 ≤ i ∧ 0 ≤ j → (i + j = 0 ↔ i = 0 ∧ j = 0) subst: [j, i] → [max_exp(i, j), max_exp(i0, j0)] flags: rewrite

Additional information

Used simplifier rules: max_exp-pos: ⊦ 0 ≤ max_exp(i, j);%max_exp-pos: ⊦ 0 ≤ max_exp(i, j); Used terms: max_exp(i, j), max_exp(i0, j0)