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Proof step 1 from proof two (conclusion)

⊦ i→n(0 -1 -1) = 2

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: int-basic1 inst: name: elim-pred seq: ⊦ i = j -1 ↔ j = i +1 subst: [j, i] → [0 -1, i] flags:

Additional information

Used simplifier rules: num-i2n-eq-01: i < 0 ⊦ i→n(i) = (* m) ↔ ~ i = n→i(m);%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%mone-02: ⊦ i + 1 = ~1 ↔ i + 1 + 1 = 0; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs% ⊦ i = j -1 ↔ j = i +1; (elim) Used terms: 0 -1, i