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Proof step 2 from proof abs-pred-02 (previous step)

0 < i ⊦ i→n(i - 1) +1 = i→n(i)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: sub: j ≤ i, 0 ≤ j ⊦ i→n(i - j) = i→n(i) - i→n(j);%num-i2n-le-02: 0 ≤ i ⊦ (* m) ≤ i→n(i) ↔ n→i(m) ≤ i;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; (s)% ⊦ i < j → (j ≤ i ↔ false); (as)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < j → ¬ j < i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ m +1 = m + 1; (s)% ⊦ 1 ≤ i ↔ 0 < i; (s)