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Proof step 21 from proof nummult-ls (previous step)

i + (i1 - 1) * i = i1 * i, 0 ≤ i + (i1 - 1) * i + j0, (i1 - 1) * i + j0 < i * i0, ~1 ≤ (i1 - 1) * i + j0, 0 < i, i0 < i1, 0 ≤ i0, j0 < i, 0 < i1, 0 ≤ j0, 0 ≤ i1 ⊦

Heuristic var elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i1, 1, i2] flags:

Additional information

Used simplifier rules: eq-dist: ⊦ (i + j) * i0 = j0 + j * i0 ↔ i * i0 = j0; from specification int-mult%one: ⊦ 1 * i = i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%numleledel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; from specification int-abs%mods-ls-04: ⊦ j0 < j ∧ 0 ≤ i ∧ 0 ≤ i0 ∧ 0 ≤ j0 → (i * j + j0 < i0 * j ↔ i < i0); from specification int-mult%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs% ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i1, 1, i2