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Proof step 19 from proof square-ex (previous step)

1 < j * j, 2 < (j + 1) * (j + 1), 0 < (j + 1) * (j + 1), 1 < (j + 1) * (j + 1), 0 < j ⊦ ∃ j0. 0 < j0 ∧ j = j0 + 1

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: j -1 abort: F discard: F computed: T

Additional information

Used simplifier rules: one-01: ⊦ i = 1 → i * j = j;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%sub-pred: ⊦ i -1 = i - 1; from specification int-abs%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%Sub-def: ⊦ i + (j - i) = j; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%bet: ⊦ 0 < i → (i ≤ 1 ↔ i = 1); from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1 Used terms: j -1